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NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers And Quadratic Equations

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MVSAT 2024

NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

The chapter Complex Number and Quadratic Equation Class 11 consists of several mathematical concepts, theorems, and formulas that are fundamental for solving sums of algebra. In this regard, NCERT Solutions for Class 11 Maths Chapter 5, prove to be extremely beneficial, as it provides step-wise solutions to all sums given in the exercises of this chapter. The NCERT Solutions for Chapter 5 Maths Class 11 is available in a free downloadable PDF version on Vedantu. These solutions are curated by senior mathematics teachers at Vedantu as per the CBSE guidelines. Download Vedantu NCERT Book Solution to get a better understanding of all the exercises questions.


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 5 - Complex Numbers and Quadratic Equations

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



Chapter at a Glance for Complex Numbers and Quadratic Equations

  • A number of the form $z-a+i b$ where $a b$ e and $i=\sqrt{-1}$ is known as a complex number, where $a$ is called the real part of $z$ and $b$ is called the imaginary part of $z$.

(i) $z$ is purely real if $\operatorname{In}(z)-0$

(ii) $z$ is purely imaginary if $\operatorname{Re}(z)-0$

  • $i=\sqrt{-1} i^2=-1, i^5=-i, i^4=1$

  • Sum of four consecutive power of iota is zero i.e, $i^n+i^{n+1}+i^{m+2}-0$, where ne $N$

  • $\sqrt{a \times b}=\sqrt{a} \times \sqrt{b}$ if atleast one of $a$ and $b$ is positive

  • $\sqrt{a \times b}=-\sqrt{a} \times \sqrt{b}$, if both $a$ and $b$ one negative

  • If $z_1=a+\vec{i}_1 z_2-c+i d$ and $z_1=z_2$ then $a+i b=c+i d$ $\Rightarrow a-c$ and $b=d$

  • $a+i b<c+i d$ or $a+i b>c+i d$ have no meaning

  • If $z_1=a+i b$ and $z_2=c+i d$ then

$$ \begin{aligned} & z_1+z_2=(a \pm c)+i(b \pm d) \\ & z_1 \times z_2-(a c-b d)+i(a d+b) \\ & \frac{z_1}{z_2}=\frac{(a c+b d)}{\left(c^2+d^2\right)}+\frac{i(b-a d)}{\left(c^2+a^d\right)} \end{aligned} $$

  • If $z=a+i b$ then $z-a-i b$

  • If $z_1 z_2$ two complex numbers, then

(i) $\overline{z_1 \pm z_2}=z_1 \pm z_2$

(ii) $\overline{z_1 \times z_2}=z_1 \times z_2$

(iii) $\left(\frac{z_1}{z_2}\right)=\frac{z_1}{z_2}$

(iv) $\overline{\left(z_1\right)}=z$

  • If $z$ is any complex number, then $|=|-\sqrt{a^2+b^2}$ and $\operatorname{ar}(z)-\tan ^{-1}\left(\frac{b}{a}\right)$

  • If $z_1 z_2$ two complex numbers, then

(i) $\left|z_1 \times z_2\right|=\left|z_1\right| F_2 \mid$

(ii) $\left|\frac{z_1}{z_2}\right|=\frac{z_1 \mid}{\left|z_2\right|}$

(iii) $\left|F_1\right|-F_2||-\left|z_1+z_2\right| \leq F_1|+| z_2 \mid$

(iv) $\operatorname{ag}\left(z_1 \times z_2\right)-\arg \left(z_1\right)+\arg \left(z_2\right)$

(v) $\operatorname{ag}\left(\frac{z_1}{z_2}\right)-\arg \left(z_1\right)-\arg \left(z_2\right)$

(vi) $\operatorname{xg}\left(z_1\right)^n-\operatorname{narg}\left(z_1\right)$

(vii) $\arg \left(\frac{z_1}{z_2}\right)--\arg \left(\frac{z_2}{z_1}\right)$

  • If $z-a+i b$ be any complex number, then polar form of $z$ is defined as $z=a+i b-|z|\{\cos \theta+i \sin \theta\}$ where $|=|=\sqrt{a^2+b^2}$ and $\theta=\tan ^{-1}\left(\frac{b}{a}\right)$

  • If $a^2+b x+c=0$ then $x=\frac{-b \pm \sqrt{D}}{a}$

Where $D$ is known as discriminant.

(i) If $D>0$, then roots are real.

(ii) If $D=0$, then roots are equal.

(iii) If $D<0$, then roots are imaginary.


Competitive Exams after 12th Science

Exercises under NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Exercise 5.1: This exercise consists of 13 questions and is focused on finding the roots of quadratic equations. The questions are based on various concepts, such as the quadratic formula, the discriminant of a quadratic equation, and finding the nature of roots. Students will learn how to find the roots of quadratic equations and analyze the nature of these roots based on the values of the discriminant.

Exercise 5.2: This exercise consists of 10 questions and is focused on solving problems related to the relationship between roots and coefficients of a quadratic equation. The questions are based on various concepts such as Vieta's formulas, the sum and product of roots, and finding the quadratic equation given its roots. Students will learn how to apply these formulas to solve problems related to quadratic equations.

Exercise 5.3: This exercise consists of 14 questions and is focused on solving problems related to complex numbers. The questions are based on various concepts such as the algebra of complex numbers, modulus and argument of a complex number, and finding the conjugate of a complex number. Students will learn how to perform operations on complex numbers and solve problems related to them.

Miscellaneous Exercise: This exercise consists of 12 questions and covers a variety of topics related to complex numbers and quadratic equations. The questions are based on various concepts such as the cube roots of unity, the geometrical representation of complex numbers, and solving problems related to the sum and product of roots of quadratic equations. This exercise will help students to revise and reinforce the concepts learned in the previous exercises.


Overall, the exercises in NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations, are designed to help students understand and apply the concepts of complex numbers and quadratic equations. The solutions to these exercises are provided in the textbook, which will help students to check their answers and understand the concepts better.


Access NCERT Solutions for Class 11 Maths Chapter 5- Complex Number and Quadratic Equations

Exercise 5.1

1. Express the given complex number in the form $ \mathrm{a+ib:}\left( \mathrm{5i} \right)\left( \mathrm{-}\dfrac{\mathrm{3}}{\mathrm{5}}\mathrm{i} \right) $ 

And evaluate

Ans:

Evaluate the complex number

$ \left( 5i \right)\left( -\dfrac{3}{5}i \right)=-5\times \dfrac{3}{5}\times i\times i $ 

$ \left( 5i \right)\left( -\dfrac{3}{5}i \right)=-3{{i}^{2}}\cdots \left[ {{i}^{2}}=-1 \right] $ 

$ \left( 5i \right)\left( -\dfrac{3}{5}i \right)=3 $ 

We get the final answer

2. Express the given complex number in the form $ {\mathrm{a + ib}}:{{\mathrm{i}}^9}{\mathrm{ + }}{{\mathrm{i}}^{19}} $ 

And evaluate

Ans:

Evaluate the complex number 

$ {{i}^{9}}+{{i}^{19}}={{i}^{4\times 2+1}}+{{i}^{4\times 4+3}} $ 

$ {{i}^{9}}+{{i}^{19}}={{\left( {{i}^{4}} \right)}^{2}}.i+{{\left( {{i}^{4}} \right)}^{4}}.{{i}^{3}}\cdots \left[ {{i}^{4}}=1,{{i}^{3}}=-1 \right] $ 

$ {{i}^{9}}+{{i}^{19}}=0 $ 

We get the final answer

3. Express the given complex number in the form $ \mathrm{a+ib:}{{\mathrm{i}}^{\mathrm{-39}}} $

And evaluate

Ans:

Evaluate the complex number 

$ {{i}^{-39}}={{i}^{4\times 9-3}} $ 

$ {{i}^{-39}}={{\left( {{i}^{4}} \right)}^{-9}}.{{i}^{-3}} $ 

$ {{i}^{-39}}=i\cdots \left[ i=-1 \right] $ 

$ {{i}^{-39}}=i $ 

We get the final answer

4. Express the given complex number in the form $ a+ib:3\left( 7+i7 \right)+i\left( 7+i7 \right) $ 

And evaluate

Ans:

Evaluate the complex number 

$ 3\left( 7+i7 \right)+i\left( 7+i7 \right)=21+21i+7i+7{{i}^{2}} $ 

$ 3\left( 7+i7 \right)+i\left( 7+i7 \right)=21+28i+7{{i}^{2}}\cdots \left[ {{i}^{2}}=-1 \right] $ 

$ 3\left( 7+i7 \right)+i\left( 7+i7 \right)=14+28i $ 

We get the final answer

5. Express the given complex number in the form

$ \mathrm{a+ib:}\left( \mathrm{1-i} \right)\mathrm{-}\left( \mathrm{-1+6i} \right) $ 

And evaluate

Ans:

Evaluate the complex number 

$ \left( 1-i \right)-\left( -1+6i \right)=1-i+1-i6 $ 

$ \left( 1-i \right)-\left( -1+6i \right)=2-7i $ 

We get the final answer

6. Express the given complex number in the form $ \mathrm{a+ib:}\left( \dfrac{\mathrm{1}}{\mathrm{5}}\mathrm{+i}\dfrac{\mathrm{2}}{\mathrm{5}} \right)\mathrm{-}\left( \mathrm{4+i}\dfrac{\mathrm{5}}{\mathrm{2}} \right) $ 

And evaluate

Ans:

Evaluate the complex number $ \left( \dfrac{\mathrm{1}}{\mathrm{5}}\mathrm{+i}\dfrac{\mathrm{2}}{\mathrm{5}} \right)\mathrm{-}\left( \mathrm{4+i}\dfrac{\mathrm{5}}{\mathrm{2}} \right)\mathrm{=}\dfrac{\mathrm{1}}{\mathrm{5}}\mathrm{+i}\dfrac{\mathrm{2}}{\mathrm{5}}\mathrm{-4-i}\dfrac{\mathrm{5}}{\mathrm{2}} $

 $ \left( \dfrac{1}{5}+i\dfrac{2}{5} \right)-\left( 4+i\dfrac{5}{2} \right)=\dfrac{-19}{5}+i\left[ \dfrac{-21}{10} \right] $ 

 $ \left( \dfrac{1}{5}+i\dfrac{2}{5} \right)-\left( 4+i\dfrac{5}{2} \right)=\dfrac{-19}{5}-\dfrac{21}{10}i $ 

We get the final answer

7. Express the given complex number in the form $ \mathrm{a+ib:}\left[ \left(\dfrac{\mathrm{1}}{\mathrm{3}}\mathrm{+i}\dfrac{\mathrm{7}}{\mathrm{3}} \right)\mathrm{+}\left( \mathrm{4+i}\dfrac{\mathrm{1}}{\mathrm{3}} \right)\mathrm{-}\left( \mathrm{-}\dfrac{\mathrm{4}}{\mathrm{3}}\mathrm{+i} \right) \right] $ 

And evaluate

Ans:

Evaluate the complex number 

$ \left[ \left( \dfrac{1}{3}+i\dfrac{7}{3} \right)+\left( 4+i\dfrac{1}{3} \right)-\left( -\dfrac{4}{3}+i \right) \right]=\dfrac{1}{3}+i\dfrac{7}{3}+4+i\dfrac{1}{3}+\dfrac{4}{3}-i $ 

$ \left[ \left( \dfrac{1}{3}+i\dfrac{7}{3} \right)+\left( 4+i\dfrac{1}{3} \right)-\left( -\dfrac{4}{3}+i \right) \right]=\left( \dfrac{1}{3}+4+\dfrac{4}{3} \right)+i\left( \dfrac{7}{3}+\dfrac{1}{3}-1 \right) $ 

$ \left[ \left( \dfrac{1}{3}+i\dfrac{7}{3} \right)+\left( 4+i\dfrac{1}{3} \right)-\left( -\dfrac{4}{3}+i \right) \right]=\dfrac{17}{3}+i\dfrac{5}{3} $ 

We get the final answer

8. Express the given complex number in the form $ \mathrm{a+ib:}{{\left( \mathrm{1-i} \right)}^{\mathrm{4}}} $ 

And evaluate

Ans:

Evaluate the complex number 

$ {{\left( 1-i \right)}^{4}}={{\left[ 1+{{i}^{2}}-2i \right]}^{2}} $ 

$ {{\left( 1-i \right)}^{4}}={{\left[ 1-1-2i \right]}^{2}} $ 

$ {{\left( 1-i \right)}^{4}}=\left( -2i \right)\times \left( -2i \right) $ 

$ {{\left( 1-i \right)}^{4}}=-4 $ 

We get the final answer

9. Express the given complex number in the form $ \mathrm{a+ib:}{{\left( \dfrac{\mathrm{1}}{\mathrm{3}}\mathrm{+3i} \right)}^{\mathrm{3}}} $ 

And evaluate

Ans:

Evaluate the complex number 

$ {{\left( \dfrac{1}{3}+3i \right)}^{3}}={{\left( \dfrac{1}{3} \right)}^{3}}+{{\left( 3i \right)}^{3}}+\dfrac{3}{3}3i\left( \dfrac{1}{3}+3i \right) $ 

$ {{\left( \dfrac{1}{3}+3i \right)}^{3}}=\dfrac{1}{27}-\left( 27i \right)+3i\left( \dfrac{1}{3}+3i \right) $  

$ {{\left( \dfrac{1}{3}+3i \right)}^{3}}=\dfrac{-242}{27}-26i $ 

We get the final answer

10. Express the given complex number in the form $ \mathrm{a+ib:}{{\left( \mathrm{-2-}\dfrac{\mathrm{1}}{\mathrm{3}}\mathrm{i} \right)}^{\mathrm{3}}} $ 

And evaluate

Ans:

Evaluate the complex number 

$ {{\left( -2-\dfrac{1}{3}i \right)}^{3}}={{\left( -1 \right)}^{3}}{{\left( 2+\dfrac{1}{3}i \right)}^{3}} $ 

$ {{\left( -2-\dfrac{1}{3}i \right)}^{3}}=-\left( {{2}^{3}}+{{\left( \dfrac{i}{3} \right)}^{3}}+6\dfrac{i}{3}\left( 2+\dfrac{i}{3} \right) \right) $  

$ {{\left( -2-\dfrac{1}{3}i \right)}^{3}}=-\dfrac{22}{3}-\dfrac{107}{27}i $ 

We get the final answer

11. Find the multiplicative inverse of the complex number

$ \mathrm{4-3i} $ 

And evaluate

Ans:

Let  $ z=4-3i $ 

Then,

$ \overline{z}=4+3i\And \left| \overline{z} \right|={{4}^{2}}+{{\left( -3 \right)}^{2}}=16+9=25 $ 

Therefore, the multiplicative inverse of  $ 4-3i $  is given by

$ {{z}^{-1}}=\dfrac{\overline{z}}{{{\left| z \right|}^{2}}}=\dfrac{4+3i}{25}=\dfrac{4}{25}+\dfrac{3}{25}i $ 

Here we got final answer

12. Find the multiplicative inverse of the complex number $ \sqrt{\mathrm{5}}\mathrm{+3i} $  

And evaluate

Ans:

Let  $ z=\sqrt{5}+3i $ 

Then,

$ \bar{z}=\sqrt{5}-3i\text{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }|z{{|}^{2}}={{(\sqrt{5})}^{2}}+{{3}^{2}}=5+9=14 $  

Therefore, the multiplicative inverse of  $ \sqrt{5}+3i $  is given by

$ {{z}^{-1}}=\dfrac{{\bar{z}}}{|z{{|}^{2}}}=\dfrac{\sqrt{5}-3i}{14}=\dfrac{\sqrt{5}}{14}-\dfrac{3i}{14} $  

Here we got final answer

13. Find the multiplicative inverse of the complex number

$ \mathrm{-i} $ 

And evaluate

Ans:

Let  $ z=-i $ 

Then,

$ \mathrm{\bar{z}=i }\!\!~\!\!\text{ and }\!\!~\!\!\text{  }\!\!|\!\!\text{ z}{{\mathrm{ }\!\!|\!\!\text{ }}^{\mathrm{2}}}\mathrm{=}{{\mathrm{1}}^{\mathrm{2}}}\mathrm{=1} $  

Therefore, the multiplicative inverse of  $ \mathrm{-i} $  is given by $ {{\mathrm{z}}^{\mathrm{-1}}}\mathrm{=}\dfrac{{\mathrm{\bar{z}}}}{\mathrm{ }\!\!|\!\!\text{ z}{{\mathrm{ }\!\!|\!\!\text{ }}^{\mathrm{2}}}}\mathrm{=}\dfrac{\mathrm{i}}{\mathrm{1}}\mathrm{=i} $  

Here we got final answer

14. Express the following expression in the form of  $ \mathrm{a+ib} $ $ \dfrac{\left( \mathrm{3+i}\sqrt{\mathrm{5}} \right)\left( \mathrm{3-i}\sqrt{\mathrm{5}} \right)}{\left( \sqrt{\mathrm{3}}\mathrm{+i}\sqrt{\mathrm{2}} \right)\mathrm{-}\left( \sqrt{\mathrm{3}}\mathrm{-i}\sqrt{\mathrm{2}} \right)} $ 

Evaluate

Ans:

The following expression $ \dfrac{\mathrm{(3+i}\sqrt{\mathrm{5}}\mathrm{)(3-i}\sqrt{\mathrm{5}}\mathrm{)}}{\mathrm{(}\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i)-(}\sqrt{\mathrm{3}}\mathrm{-i}\sqrt{\mathrm{2}}\mathrm{)}}\mathrm{=}\dfrac{{{\mathrm{(3)}}^{\mathrm{2}}}\mathrm{-(i}\sqrt{\mathrm{5}}{{\mathrm{)}}^{\mathrm{2}}}}{\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i-}\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i}} $ 

 $ \begin{align} &\dfrac{\mathrm{(3+i}\sqrt{\mathrm{5}}\mathrm{)(3-i}\sqrt{\mathrm{5}}\mathrm{)}}{\mathrm{(}\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i)-(}\sqrt{\mathrm{3}}\mathrm{-i}\sqrt{\mathrm{2}}\mathrm{)}}\mathrm{=}\dfrac{\mathrm{9-5}{{\mathrm{i}}^{\mathrm{2}}}}{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}} \\ &\dfrac{\mathrm{(3+i}\sqrt{\mathrm{5}}\mathrm{)(3-i}\sqrt{\mathrm{5}}\mathrm{)}}{\mathrm{(}\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i)-(}\sqrt{\mathrm{3}}\mathrm{-i}\sqrt{\mathrm{2}}\mathrm{)}}\mathrm{=}\dfrac{\mathrm{9-5}\left( \mathrm{-1} \right)}{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{i}} \\ &\dfrac{\mathrm{(3+i}\sqrt{\mathrm{5}}\mathrm{)(3-i}\sqrt{\mathrm{5}}\mathrm{)}}{\mathrm{(}\sqrt{\mathrm{3}}\mathrm{+}\sqrt{\mathrm{2}}\mathrm{i)-(}\sqrt{\mathrm{3}}\mathrm{-i}\sqrt{\mathrm{2}}\mathrm{)}}\mathrm{=}\dfrac{\mathrm{-7}\sqrt{\mathrm{2i}}}{\mathrm{2}} \\  \end{align} $  

Here we got final answer

Exercise 5.2

1. Find the modulus and the argument of the complex number $ \mathrm{z=-1-i}\sqrt{\mathrm{3}} $

Evaluate

Ans:

The complex number is

$ \mathrm{z=-1-i}\sqrt{\mathrm{3}} $  

Let  $ \mathrm{rcos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ rsin }\!\!\theta\!\!\text{ =-}\sqrt{\mathrm{3}} $ 

Squaring and adding

$ {{\mathrm{(rcos }\!\!\theta\!\!\text{ )}}^{\mathrm{2}}}\mathrm{+(rsin }\!\!\theta\!\!\text{ }{{\mathrm{)}}^{\mathrm{2}}}\mathrm{=(-1}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{+(-}\sqrt{\mathrm{3}}{{\mathrm{)}}^{\mathrm{2}}} $ 

$ \begin{matrix} {{\mathrm{r}}^{\mathrm{2}}}\left( \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ } \right)\mathrm{=1+3}  \\ {{\mathrm{r}}^{\mathrm{2}}}\mathrm{=4}\quad \left[ \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ =1} \right]  \\ \end{matrix} $ 

$ \mathrm{r=}\sqrt{\mathrm{4}}\mathrm{=2}\quad \mathrm{ }\!\![\!\!\text{  }\!\!~\!\!\text{ Conventionally, }\!\!~\!\!\text{ r0 }\!\!]\!\!\text{ } $ 

$ \begin{matrix} \mathrm{Modulus }\!\!~\!\!\text{ =2}  \\ \mathrm{2cos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ 2sin }\!\!\theta\!\!\text{ =-}\sqrt{\mathrm{3}}  \\ \mathrm{cos }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{-1}}{\mathrm{2}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ sin }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{-}\sqrt{\mathrm{3}}}{\mathrm{2}}  \\ \end{matrix} $  

Since both the values of  $ \mathrm{sin }\!\!\theta\!\!\text{  }\!\!~\!\!\text{ and }\!\!~\!\!\text{ cos }\!\!\theta\!\!\text{ } $  negative and  $ \mathrm{sin }\!\!\theta\!\!\text{  }\!\!~\!\!\text{ and }\!\!~\!\!\text{ cos }\!\!\theta\!\!\text{ } $  are negative in 3rd quadrant,

$ \mathrm{ }\!\!~\!\!\text{ Argument }\!\!~\!\!\text{ =-}\left( \mathrm{ }\!\!\pi\!\!\text{ -}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{3}} \right)\mathrm{=}\dfrac{\mathrm{-2 }\!\!\pi\!\!\text{ }}{\mathrm{3}} $ 

Thus, the modulus and argument of the complex number  $ \mathrm{-1-}\sqrt{\mathrm{3}}\mathrm{i }\!\!~\!\!\text{ are }\!\!~\!\!\text{ 2 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ -}\dfrac{\mathrm{2 }\!\!\pi\!\!\text{ }}{\mathrm{3}} $ 

Respectively

2. Find the modulus and the argument of the complex number

$ \mathrm{z=-}\sqrt{\mathrm{3}}\mathrm{+i} $ 

Evaluate

Ans:

The complex number is

 $ \mathrm{z=-}\sqrt{\mathrm{3}}\mathrm{+i} $  

Let  $ \mathrm{rcos }\!\!\theta\!\!\text{ =-}\sqrt{\mathrm{3}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ rsin }\!\!\theta\!\!\text{ =1} $ 

squaring and adding

$ {{\mathrm{(rcos }\!\!\theta\!\!\text{ )}}^{\mathrm{2}}}\mathrm{+(rsin }\!\!\theta\!\!\text{ }{{\mathrm{)}}^{\mathrm{2}}}\mathrm{=(-}\sqrt{\mathrm{3}}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{+(-1}{{\mathrm{)}}^{\mathrm{2}}} $ 

$ \begin{matrix} {{\mathrm{r}}^{\mathrm{2}}}\mathrm{=3+1=4LLL}\left[ \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ =1} \right]  \\ \mathrm{r=}\sqrt{\mathrm{4}}\mathrm{=2LLL }\!\![\!\!\text{  }\!\!~\!\!\text{ Conventionally, }\!\!~\!\!\text{ r0 }\!\!]\!\!\text{ }  \\ \end{matrix} $  

$ \begin{matrix} \mathrm{Modulus }\!\!~\!\!\text{ =2}  \\ \mathrm{2cos }\!\!\theta\!\!\text{ =-}\sqrt{\mathrm{3}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ 2sin }\!\!\theta\!\!\text{ =1}  \\ \end{matrix} $  

$ \begin{align} & \mathrm{cos }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{-}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ sin }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{1}}{\mathrm{2}} \\  & \mathrm{ }\!\!\theta\!\!\text{ = }\!\!\pi\!\!\text{ -}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{6}}\mathrm{=}\dfrac{\mathrm{5 }\!\!\pi\!\!\text{ }}{\mathrm{6}}\mathrm{LL }\!\![\!\!\text{ As }\!\!~\!\!\text{  }\!\!\theta\!\!\text{  }\!\!~\!\!\text{ lies in the II quadrant }\!\!]\!\!\text{ } \\ \end{align} $ 

Thus, the modulus and argument of the complex number  $ \mathrm{-}\sqrt{\mathrm{3}}\mathrm{+i }\!\!~\!\!\text{ are }\!\!~\!\!\text{ 2 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\dfrac{\mathrm{5 }\!\!\pi\!\!\text{ }}{\mathrm{6}} $  

Respectively

3. Convert the given complex number in polar form

$ \mathrm{1-i} $ 

And evaluate

Ans:

The complex number is

$ 1-i $ 

Let  $ r\cos \theta =1\text{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }r\sin \theta =-1 $ 

squaring and adding

$ \begin{matrix} {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta ={{1}^{2}}+{{(-1)}^{2}}  \\ \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta  \right)=1+1  \\ \end{matrix} $ 

$ \begin{align} & {{r}^{2}}=2 \\  & r=\sqrt{2}\quad [\text{ }\!\!~\!\!\text{ Conventionally, }\!\!~\!\!\text{ }r>0] \\  \end{align} $ 

$ \begin{align} & \sqrt{\mathrm{2}}\mathrm{cos }\!\!\theta\!\!\text{ =1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{sin }\!\!\theta\!\!\text{ =-1} \\ & \mathrm{cos }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ sin }\!\!\theta\!\!\text{ =-}\dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\  & \mathrm{ }\!\!\theta\!\!\text{ =-}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{4}}\quad \mathrm{ }\!\![\!\!\text{  }\!\!~\!\!\text{ As }\!\!~\!\!\text{  }\!\!\theta\!\!\text{  }\!\!~\!\!\text{ liesin the IV quadrant }\!\!~\!\!\text{  }\!\!]\!\!\text{ } \\ \end{align} $ 

$ \mathrm{ }\!\!\!\!\text{ 1-i=rcos }\!\!\theta\!\!\text{ +irsin }\!\!\theta\!\!\text{ =}\sqrt{\mathrm{2}}\mathrm{cos}\left( \mathrm{-}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{4}} \right)\mathrm{+i}\sqrt{\mathrm{2}}\mathrm{sin}\left( \mathrm{-}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{4}} \right)\mathrm{=}\sqrt{\mathrm{2}}\left[ \mathrm{cos}\left( \mathrm{-}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{4}} \right)\mathrm{+isin}\left( \mathrm{-}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{4}} \right) \right] $ 

Required polar form

4. Convert the given complex number in polar form

$ \mathrm{-1+i} $ 

And evaluate

Ans:

The complex number is

$ \mathrm{-1+i} $ 

Let  $ \mathrm{rcos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ rsin }\!\!\theta\!\!\text{ =1} $ 

Squaring and adding

$ \begin{align} & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +}{{\mathrm{r}}^{\mathrm{2}}}\mathrm{si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ =(-1}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}} \\ & {{\mathrm{r}}^{\mathrm{2}}}\left( \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ } \right)\mathrm{=1+1} \\  & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{=2} \\  & \mathrm{r=}\sqrt{\mathrm{2}} \\  \end{align} $ 

$ \sqrt{\mathrm{2}}\mathrm{cos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{sin }\!\!\theta\!\!\text{ =1} $ 

$ \begin{align} & \mathrm{cos }\!\!\theta\!\!\text{ =-}\dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{sin }\!\!\theta\!\!\text{ =1} \\  & \mathrm{ }\!\!\!\!\text{  }\!\!\theta\!\!\text{ = }\!\!\pi\!\!\text{ -}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{=}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{L}\left[ \mathrm{As }\!\!~\!\!\text{  }\!\!\theta\!\!\text{  }\!\!~\!\!\text{ lies in the II quadrant} \right] \\ \end{align} $ 

It can be written,

$ \mathrm{ }\!\!\!\!\text{ -1+i=rcos }\!\!\theta\!\!\text{ +irsin }\!\!\theta\!\!\text{ =}\sqrt{\mathrm{2}}\mathrm{cos}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{+i}\sqrt{\mathrm{2}}\mathrm{sin}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{=}\sqrt{\mathrm{2}}\left( \mathrm{cos}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{+isin}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}} \right) $ 

Required polar form

 5. Convert the given complex number in polar form $ \mathrm{-1-i} $ 

And evaluate

Ans:

The complex number is

$ \mathrm{-1-i} $ 

Let  $ \mathrm{rcos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ rsin }\!\!\theta\!\!\text{ =-1} $ 

Squaring and adding

$ \begin{align} & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +}{{\mathrm{r}}^{\mathrm{2}}}\mathrm{si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ =(-1}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{+(-1}{{\mathrm{)}}^{\mathrm{2}}} \\ & {{\mathrm{r}}^{\mathrm{2}}}\left( \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ } \right)\mathrm{=1+1} \\ & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{=2} \\ & \mathrm{r=}\sqrt{\mathrm{2}} \\ \end{align} $ 

$ \begin{align} & \mathrm{ }\!\!\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{cos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{sin }\!\!\theta\!\!\text{ =-1} \\  & \mathrm{cos }\!\!\theta\!\!\text{ =-}\dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ sin }\!\!\theta\!\!\text{ =-}\dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\  & \mathrm{ }\!\!\!\!\text{  }\!\!\theta\!\!\text{ --}\left( \mathrm{ }\!\!\pi\!\!\text{ -}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{4}} \right)\mathrm{--}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\quad \mathrm{ }\!\!~\!\!\text{ }\left[ \mathrm{As }\!\!~\!\!\text{ 0 }\!\!~\!\!\text{ lies in the III quadrant} \right] \\ \end{align} $ 

$ \mathrm{ }\!\!\!\!\text{ -1-i=rcos }\!\!\theta\!\!\text{ +irsin }\!\!\theta\!\!\text{ =}\sqrt{\mathrm{2}}\mathrm{cos}\dfrac{\mathrm{-3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{+i}\sqrt{\mathrm{2}}\mathrm{sin}\dfrac{\mathrm{-3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{=}\sqrt{\mathrm{2}}\left( \mathrm{cos}\dfrac{\mathrm{-3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{+isin}\dfrac{\mathrm{-3 }\!\!\pi\!\!\text{ }}{\mathrm{4}} \right) $ 

Required polar form

6. Convert the given complex number in polar form

$ \mathrm{-3} $ 

And evaluate

Ans:

The complex number is

$ \mathrm{-3} $ 

Let  $ \mathrm{rcos }\!\!\theta\!\!\text{ =-3 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ rsin }\!\!\theta\!\!\text{ =0} $ 

Squaring and adding

$ \begin{align} & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +}{{\mathrm{r}}^{\mathrm{2}}}\mathrm{si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ =(-3}{{\mathrm{)}}^{\mathrm{2}}} \\  & {{\mathrm{r}}^{\mathrm{2}}}\left( \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ } \right)\mathrm{=9} \\ \end{align} $ 

$ \begin{align} & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{=9} \\  & \mathrm{r=}\sqrt{\mathrm{9}}\mathrm{=3} \\ \end{align} $ 

$ \begin{align} & \mathrm{ }\!\!\!\!\text{ 3cos }\!\!\theta\!\!\text{ =-3 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ 3sin }\!\!\theta\!\!\text{ =0} \\ & \mathrm{cos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ sin=0} \\ & \mathrm{ }\!\!\!\!\text{  }\!\!\theta\!\!\text{ = }\!\!\pi\!\!\text{ } \\ \end{align} $ 

$ \mathrm{ }\!\!\!\!\text{ -3=rcos }\!\!\theta\!\!\text{ +irsin }\!\!\theta\!\!\text{ =3cos }\!\!\pi\!\!\text{ +i3sin }\!\!\pi\!\!\text{ =3(cos }\!\!\pi\!\!\text{ +isin }\!\!\pi\!\!\text{ )} $ 

Required polar form

 

7. Convert the given complex number in polar form $ \sqrt{\mathrm{3}}\mathrm{+i} $ 

And evaluate

Ans:

The complex number is

$ \sqrt{\mathrm{3}}\mathrm{+i} $ 

Let  $ \mathrm{rcos }\!\!\theta\!\!\text{ =}\sqrt{\mathrm{3}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ rsin }\!\!\theta\!\!\text{ =1} $ 

Squaring and adding

$ \begin{align} & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +}{{\mathrm{r}}^{\mathrm{2}}}\mathrm{si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ =(}\sqrt{\mathrm{3}}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}} \\ & {{\mathrm{r}}^{\mathrm{2}}}\left( \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ } \right)\mathrm{=3+1} \\ & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{=4} \\  & \mathrm{r=}\sqrt{\mathrm{4}}\mathrm{=2} \\ \end{align} $ 

$ \begin{align} & \mathrm{2cos }\!\!\theta\!\!\text{ =}\sqrt{\mathrm{3}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ 2sin }\!\!\theta\!\!\text{ =1} \\ & \mathrm{cos }\!\!\theta\!\!\text{ =}\dfrac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ sin }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{1}}{\mathrm{2}} \\ & \mathrm{ }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{6}}\quad \mathrm{ }\!\!~\!\!\text{ }\left[ \mathrm{As }\!\!~\!\!\text{  }\!\!\theta\!\!\text{  }\!\!~\!\!\text{ lies in the I quadrant} \right] \\ \end{align} $ 

$ \mathrm{ }\!\!\!\!\text{ }\sqrt{\mathrm{3}}\mathrm{+i=rcos }\!\!\theta\!\!\text{ +irsin }\!\!\theta\!\!\text{ =2cos}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{6}}\mathrm{+i2sin}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{6}}\mathrm{=2}\left( \mathrm{cos}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{6}}\mathrm{+isin}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{6}} \right) $ 

Required polar form


8. Convert the given complex number in polar form

$ \mathrm{i} $ 

And evaluate

Ans:

The complex number is

$ \mathrm{i} $ 

Let  $ \mathrm{rcos }\!\!\theta\!\!\text{ =0 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ rsin }\!\!\theta\!\!\text{ =1} $ 

Squaring and adding

$ \begin{align} & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +}{{\mathrm{r}}^{\mathrm{2}}}\mathrm{si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ =}{{\mathrm{0}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}} \\  & {{\mathrm{r}}^{\mathrm{2}}}\left( \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ } \right)\mathrm{=1} \\ \end{align} $ 

$ \begin{align} & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{=1} \\  & \mathrm{r=}\sqrt{\mathrm{1}}\mathrm{=1}\quad \mathrm{ }\!\![\!\!\text{  }\!\!~\!\!\text{ Conventionally, }\!\!~\!\!\text{ r0 }\!\!]\!\!\text{ } \\ \end{align} $ 

$ \begin{align} & \mathrm{cos }\!\!\theta\!\!\text{ =0 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ sin }\!\!\theta\!\!\text{ =1} \\ & \mathrm{ }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{2}} \\  & \mathrm{i=rcos }\!\!\theta\!\!\text{ +irsin }\!\!\theta\!\!\text{ =cos}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{2}}\mathrm{+isin}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{2}} \\ \end{align} $ 

Required polar form

Exercise 5.3

1. Solve the equation
$ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+3=0} $  

And evaluate

Ans:

Quadratic equation  $ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+3=0} $ 

General form  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

We obtain  $ \mathrm{a=1,b=0, }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=3} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\mathrm{0}}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ 1 }\!\!\times\!\!\text{ 3=-12} $ 

Therefore, the required solutions are

$ \begin{align}& \mathrm{=}\dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{ }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-12}}}{\mathrm{2 }\!\!\times\!\!\text{ 1}}\mathrm{=}\dfrac{\mathrm{ }\!\!\pm\!\!\text{ }\sqrt{\mathrm{12}}\mathrm{i}}{\mathrm{2}} \\ & \mathrm{=}\dfrac{\mathrm{ }\!\!\pm\!\!\text{ 2}\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{2}}\mathrm{= }\!\!\pm\!\!\text{ }\sqrt{\mathrm{3}}\mathrm{i} \\ \end{align} $

 

2. Solve the equation
$ \mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+x+1=0} $  

And evaluate

Ans:

Quadratic equation  $ \mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+x+1=0} $ 

General form  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

We obtain  $ \mathrm{a=2,b=1, }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=1} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\mathrm{1}}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 1=-7} $ 

Therefore, the required solutions are

$ \mathrm{=}\dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{ }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-7}}}{\mathrm{2 }\!\!\times\!\!\text{ 2}}\mathrm{=}\dfrac{\mathrm{ }\!\!\pm\!\!\text{ }\sqrt{\mathrm{7}}\mathrm{i}}{\mathrm{4}} $ 


3. Solve the equation
$ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+3x+9=0} $  

And evaluate

Ans:

Quadratic equation  $ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+3x+9=0} $ 

General form  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

We obtain  $ \mathrm{a=1,b=3, }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=9} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\mathrm{3}}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ 1 }\!\!\times\!\!\text{ 9=-27} $ 

Therefore, the required solutions are

$ \mathrm{=}\dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-3 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-27}}}{\mathrm{2 }\!\!\times\!\!\text{ 1}}\mathrm{=}\dfrac{\mathrm{-3 }\!\!\pm\!\!\text{ 3}\sqrt{\mathrm{3}}\mathrm{i}}{\mathrm{2}} $


4. Solve the equation
$ \mathrm{-}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+x-2=0} $  

And evaluate

Ans:

Quadratic equation  $ \mathrm{-}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+x-2=0} $ 

General form  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

We obtain  $ \mathrm{a=-1,b=1, }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=-2} $ 

Therefore, the discriminant of the given equation is

 $ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\mathrm{1}}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ -1 }\!\!\times\!\!\text{ -2=-7} $ 

Therefore, the required solutions are

$ \mathrm{=}\dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-1 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-7}}}{\mathrm{2 }\!\!\times\!\!\text{ -1}}\mathrm{=}\dfrac{\mathrm{-1 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{7}}\mathrm{i}}{\mathrm{-2}} $ 


5. Solve the equation
$ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+3x+5=0} $  

And evaluate

Ans:

Quadratic equation  $ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+3x+5=0} $ 

General form  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

We obtain  $ \mathrm{a=1,b=3, }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=5} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\mathrm{3}}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ 1 }\!\!\times\!\!\text{ 5=-11} $ 

Therefore, the required solutions are

$ \mathrm{=}\dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-3 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-11}}}{\mathrm{2 }\!\!\times\!\!\text{ 1}}\mathrm{=}\dfrac{\mathrm{-3 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{11}}\mathrm{i}}{\mathrm{2}} $

6. Solve the equation
$ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-x+2=0} $  

And evaluate

Ans:

Quadratic equation  $ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-x+2=0} $ 

General form  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

We obtain  $ \mathrm{a=1,b=3-1, }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=2} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\left( \mathrm{-1} \right)}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ 1 }\!\!\times\!\!\text{ 2=-7} $ 

Therefore, the required solutions are

$ \mathrm{=}\dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-}\left( \mathrm{-1} \right)\mathrm{ }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-7}}}{\mathrm{2 }\!\!\times\!\!\text{ 1}}\mathrm{=}\dfrac{\mathrm{1 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{7}}\mathrm{i}}{\mathrm{2}} $

7. Solve the equation
$\sqrt{\mathrm{2}}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+x+}\sqrt{\mathrm{2}}\mathrm{=0} $  

And evaluate

Ans:

Quadratic equation  $ \sqrt{\mathrm{2}}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+x+}\sqrt{\mathrm{2}}\mathrm{=0} $ 

General form  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

We obtain  $ \mathrm{a=}\sqrt{\mathrm{2}}\mathrm{,b=1, }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=}\sqrt{\mathrm{2}} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\mathrm{1}}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{ }\!\!\times\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{=-7} $ 

Therefore, the required solutions are

$ \mathrm{=}\dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-1 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-7}}}{\mathrm{2 }\!\!\times\!\!\text{ }\sqrt{\mathrm{2}}}\mathrm{=}\dfrac{\mathrm{-1 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{7}}\mathrm{i}}{\mathrm{2}\sqrt{\mathrm{2}}} $

8. Solve the equation
$\sqrt{\mathrm{3}}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}\sqrt{\mathrm{2}}\mathrm{x+3}\sqrt{\mathrm{3}}\mathrm{=0} $  

And evaluate

Ans:

Quadratic equation  $ \sqrt{\mathrm{3}}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}\sqrt{\mathrm{2}}\mathrm{x+3}\sqrt{\mathrm{3}}\mathrm{=0} $ 

General form  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

We obtain  $ \mathrm{a=}\sqrt{\mathrm{3}}\mathrm{,b=-}\sqrt{\mathrm{2}}\mathrm{, }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=3}\sqrt{\mathrm{3}} $ 

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\left( \mathrm{-}\sqrt{\mathrm{2}} \right)}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ }\sqrt{\mathrm{3}}\mathrm{ }\!\!\times\!\!\text{ 3}\sqrt{\mathrm{3}}\mathrm{=-34} $  $ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\left( \mathrm{-}\sqrt{\mathrm{2}} \right)}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ }\sqrt{\mathrm{3}}\mathrm{ }\!\!\times\!\!\text{ 3}\sqrt{\mathrm{3}}\mathrm{=-34} $ 

Therefore, the required solutions are

$ \mathrm{=}\dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-}\left( \mathrm{-}\sqrt{\mathrm{2}} \right)\mathrm{ }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-34}}}{\mathrm{2 }\!\!\times\!\!\text{ }\sqrt{\mathrm{3}}}\mathrm{=}\dfrac{\sqrt{\mathrm{2}}\mathrm{ }\!\!\pm\!\!\text{ }\sqrt{\mathrm{34}}\mathrm{i}}{\mathrm{2}\sqrt{\mathrm{3}}} $


9. Solve the equation
${{\mathrm{x}}^{\mathrm{2}}}\mathrm{+x+}\dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{=0} $  

And evaluate

Ans:

Quadratic equation  $ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+x+}\dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}}\mathrm{=0} $

General form  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

We obtain  $ \mathrm{a=}\sqrt{\mathrm{2}}\mathrm{,b=}\sqrt{\mathrm{2}}\mathrm{, }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=1} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\left( \sqrt{\mathrm{2}} \right)}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{ }\!\!\times\!\!\text{ 1=2-4}\sqrt{\mathrm{2}} $ 

Therefore, the required solutions are

$ \mathrm{=}\dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-}\left( \sqrt{\mathrm{2}} \right)\mathrm{ }\!\!\pm\!\!\text{ }\sqrt{\mathrm{2-4}\sqrt{\mathrm{2}}}}{\mathrm{2 }\!\!\times\!\!\text{ }\sqrt{\mathrm{2}}}\mathrm{=}\dfrac{\mathrm{-1 }\!\!\pm\!\!\text{ }\left( \sqrt{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{-1}} \right)\mathrm{i}}{\mathrm{2}} $


10. Solve the equation
${{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}\dfrac{\mathrm{x}}{\sqrt{\mathrm{2}}}\mathrm{+1=0} $  

And evaluate

Ans:

Quadratic equation  $ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}\dfrac{\mathrm{x}}{\sqrt{\mathrm{2}}}\mathrm{+1=0} $ 

General form  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

We obtain  $ \mathrm{a=}\sqrt{\mathrm{2}}\mathrm{,b=1, }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=}\sqrt{\mathrm{2}} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=}{{\left( \mathrm{1} \right)}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{ }\!\!\times\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{=-7} $ 

Therefore, the required solutions are

$ \mathrm{=}\dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-}\left( \mathrm{1} \right)\mathrm{ }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-7}}}{\mathrm{2 }\!\!\times\!\!\text{ }\sqrt{\mathrm{2}}}\mathrm{=}\dfrac{\mathrm{-1 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{7}}\mathrm{i}}{\mathrm{2}\sqrt{\mathrm{2}}} $


Miscellaneous Exercise

1. Evaluate
$ {{\left[ {{\mathrm{i}}^{\mathrm{18}}}\mathrm{+}{{\left( \dfrac{\mathrm{1}}{\mathrm{i}} \right)}^{\mathrm{25}}} \right]}^{\mathrm{3}}} $  

The expression 

Ans:

Expression

$ {{\left[ {{\mathrm{i}}^{\mathrm{18}}}\mathrm{+}{{\left( \dfrac{\mathrm{1}}{\mathrm{i}} \right)}^{\mathrm{25}}} \right]}^{\mathrm{3}}}\mathrm{=}{{\left[ {{\mathrm{i}}^{\mathrm{4 }\!\!\times\!\!\text{ 4+2}}}\mathrm{+}\dfrac{\mathrm{1}}{{{\mathrm{i}}^{\mathrm{4 }\!\!\times\!\!\text{ 6+1}}}} \right]}^{\mathrm{3}}} $ 

$ \begin{align} & \mathrm{=}{{\left[ {{\left( {{\mathrm{i}}^{\mathrm{4}}} \right)}^{\mathrm{4}}}\mathrm{ }\!\!\times\!\!\text{ }{{\mathrm{i}}^{\mathrm{2}}}\mathrm{+}\dfrac{\mathrm{1}}{{{\left( {{\mathrm{i}}^{\mathrm{4}}} \right)}^{\mathrm{6}}}\mathrm{ }\!\!\times\!\!\text{ i}} \right]}^{\mathrm{3}}} \\  & \mathrm{=}{{\left[ {{\mathrm{i}}^{\mathrm{2}}}\mathrm{+}\dfrac{\mathrm{1}}{\mathrm{i}} \right]}^{\mathrm{3}}}\quad \left[ {{\mathrm{i}}^{\mathrm{4}}}\mathrm{=1} \right] \\ & \mathrm{=}{{\left[ \mathrm{-1+}\dfrac{\mathrm{1}}{\mathrm{i}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{i}}{\mathrm{i}} \right]}^{\mathrm{3}}}\quad \left[ {{\mathrm{i}}^{\mathrm{2}}}\mathrm{=-1} \right] \\ & \mathrm{=}{{\left[ \mathrm{-1+}\dfrac{\mathrm{i}}{{{\mathrm{i}}^{\mathrm{2}}}} \right]}^{\mathrm{3}}} \\ \end{align} $ 

$ \begin{align} & \mathrm{= }\!\![\!\!\text{ -1-i}{{\mathrm{ }\!\!]\!\!\text{ }}^{\mathrm{3}}} \\ & \mathrm{=(-1}{{\mathrm{)}}^{\mathrm{3}}}{{\mathrm{ }\!\![\!\!\text{ 1+i }\!\!]\!\!\text{ }}^{\mathrm{3}}} \\  & \mathrm{=-}\left[ {{\mathrm{1}}^{\mathrm{3}}}\mathrm{+}{{\mathrm{i}}^{\mathrm{3}}}\mathrm{+3 }\!\!\times\!\!\text{ 1 }\!\!\times\!\!\text{ i(1+i)} \right] \\  & \mathrm{=-}\left[ \mathrm{1+}{{\mathrm{i}}^{\mathrm{3}}}\mathrm{+3i+3}{{\mathrm{i}}^{\mathrm{2}}} \right] \\ & \mathrm{=- }\!\![\!\!\text{ 1-i+3i-3 }\!\!]\!\!\text{ } \\  & \mathrm{=- }\!\![\!\!\text{ -2+2i }\!\!]\!\!\text{ } \\  & \mathrm{=2-2i} \\ \end{align} $ 

The expression is evaluated

2. For any two complex numbers  $ {{\mathrm{z}}_{\mathrm{1}}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }{{\mathrm{z}}_{\mathrm{2}}} $ , prove that $ \mathrm{Re}\left( {{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}} \right)\mathrm{=Re}{{\mathrm{z}}_{\mathrm{1}}}\mathrm{Re}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{-Im}{{\mathrm{z}}_{\mathrm{1}}}\mathrm{Im}{{\mathrm{z}}_{\mathrm{2}}} $  

Ans:

Let $ {{\mathrm{z}}_{\mathrm{1}}}\mathrm{=}{{\mathrm{x}}_{\mathrm{1}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{1}}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{2}}} $ 

 $ \begin{matrix} \mathrm{ }\!\!\!\!\text{ }{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=}\left( {{\mathrm{x}}_{\mathrm{1}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{1}}} \right)\left( {{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{2}}} \right)  \\ \mathrm{=}{{\mathrm{x}}_{\mathrm{1}}}\left( {{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{2}}} \right)\mathrm{+i}{{\mathrm{y}}_{\mathrm{1}}}\left( {{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{2}}} \right)  \\ \mathrm{=}{{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{+}{{\mathrm{i}}^{\mathrm{2}}}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}}  \\ \end{matrix} $ 

 $ \begin{align} & \mathrm{=}{{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}}\mathrm{+i}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}} \\ & \mathrm{=}\left( {{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}} \right)\mathrm{+i}\left( {{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}} \right) \\  & \mathrm{Re}\left( {{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}} \right)\mathrm{=}{{\mathrm{x}}_{\mathrm{1}}}{{\mathrm{x}}_{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}_{\mathrm{1}}}{{\mathrm{y}}_{\mathrm{2}}} \\  & \mathrm{Re}\left( {{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}} \right)\mathrm{=Re}{{\mathrm{z}}_{\mathrm{1}}}\mathrm{Re}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{-Im}{{\mathrm{z}}_{\mathrm{1}}}\mathrm{Im}{{\mathrm{z}}_{\mathrm{2}}} \\  \end{align} $ 

Hence, proved

3. Reduce  $ \left( \dfrac{\mathrm{1}}{\mathrm{1-4i}}\mathrm{-}\dfrac{\mathrm{2}}{\mathrm{1+i}} \right)\left( \dfrac{\mathrm{3-4i}}{\mathrm{5+i}} \right) $  to the standard form 

Ans:

Expression 

$ \begin{align} & \left( \dfrac{\mathrm{1}}{\mathrm{1-4i}}\mathrm{-}\dfrac{\mathrm{2}}{\mathrm{1+i}} \right)\left( \dfrac{\mathrm{3-4i}}{\mathrm{5+i}} \right)\mathrm{=}\left[ \dfrac{\mathrm{(1+i)-2(1-4i)}}{\mathrm{(1-4i)(1+i)}} \right]\left[ \dfrac{\mathrm{3-4i}}{\mathrm{5+i}} \right] \\  & \mathrm{=}\left[ \dfrac{\mathrm{1+i-2+8i}}{\mathrm{1+i-4i-4}{{\mathrm{i}}^{\mathrm{2}}}} \right]\left[ \dfrac{\mathrm{3-4i}}{\mathrm{5+i}} \right]\mathrm{=}\left[ \dfrac{\mathrm{-1+9i}}{\mathrm{5-3i}} \right]\left[ \dfrac{\mathrm{3-4i}}{\mathrm{5+i}} \right] \\  & \mathrm{=}\left[ \dfrac{\mathrm{-3+4i+27i-36}{{\mathrm{i}}^{\mathrm{2}}}}{\mathrm{25+5i-15i-3}{{\mathrm{i}}^{\mathrm{2}}}} \right]\mathrm{=}\dfrac{\mathrm{33+31i}}{\mathrm{28-10i}}\mathrm{=}\dfrac{\mathrm{33+31i}}{\mathrm{2(14-5i)}} \\  \end{align} $ 

$ \begin{align} & \mathrm{=}\dfrac{\mathrm{(33+31i)}}{\mathrm{2(14-5i)}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{(14+5i)}}{\mathrm{(14+5i)}}\mathrm{ }\!\!~\!\!\text{  }\!\![\!\!\text{ On multiplying numerator and denominator by(14+5i) }\!\!]\!\!\text{ } \\ &\mathrm{=}\dfrac{\mathrm{462+165i+434i+155}{{\mathrm{i}}^{\mathrm{2}}}}{\mathrm{2}\left[ {{\mathrm{(14)}}^{\mathrm{2}}}\mathrm{-(5i}{{\mathrm{)}}^{\mathrm{2}}} \right]}\mathrm{=}\dfrac{\mathrm{307+599i}}{\mathrm{2}\left( \mathrm{196-25}{{\mathrm{i}}^{\mathrm{2}}} \right)} \\ &\mathrm{=}\dfrac{\mathrm{307+599i}}{\mathrm{2(221)}}\mathrm{=}\dfrac{\mathrm{307+599i}}{\mathrm{442}}\mathrm{=}\dfrac{\mathrm{307}}{\mathrm{442}}\mathrm{+}\dfrac{\mathrm{599i}}{\mathrm{442}} \\ \end{align} $ 

This is the required standard form


4. If  $ \mathrm{x-iy=}\sqrt{\dfrac{\mathrm{a-ib}}{\mathrm{c-id}}} $  prove that  $ {{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}\mathrm{=}\dfrac{{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} $  

Ans:

Expression 

$ \begin{align} & \mathrm{x-iy=}\sqrt{\dfrac{\mathrm{a-ib}}{\mathrm{c-id}}} \\  & \left. \mathrm{=}\sqrt{\dfrac{\mathrm{a-ib}}{\mathrm{c-id}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{c+id}}{\mathrm{c+id}}}\quad \mathrm{ }\!\!~\!\!\text{  }\!\![\!\!\text{ On multiplying numerator and denominator by }\!\!~\!\!\text{ (c+id)} \right] \\  &\mathrm{=}\sqrt{\dfrac{\mathrm{(ac+bd)+i(ad-bc)}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}}} \\ \end{align} $  

$ \begin{align} & \mathrm{ }\!\!\!\!\text{ (x-iy}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{=}\dfrac{\mathrm{(ac+bd)+i(ad-bc)}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} \\ &{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{-2ixy=}\dfrac{\mathrm{(ac+bd)+i(ad-bc)}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} \\ \end{align} $ 

On comparing

$ \begin{align} &{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{=}\dfrac{\mathrm{ac+bd}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}}\mathrm{,-2xy=}\dfrac{\mathrm{ad-bc}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}}\mathrm{}...\mathrm{(1)} \\  & {{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}\mathrm{=}{{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}\mathrm{+4}{{\mathrm{x}}^{\mathrm{2}}}{{\mathrm{y}}^{\mathrm{2}}} \\  & \mathrm{=}{{\left( \dfrac{\mathrm{ac+bd}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} \right)}^{\mathrm{2}}}\mathrm{+}\left( \dfrac{\mathrm{ad-bc}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} \right) \\ &\mathrm{=}\dfrac{{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{d}}^{\mathrm{2}}}\mathrm{+2acbd+}{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{d}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{c}}^{\mathrm{2}}}\mathrm{-2adbc}}{{{\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}} \\  \end{align} $ 

$ \begin{align} &\mathrm{=}\dfrac{{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{d}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{d}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{c}}^{\mathrm{2}}}}{{{\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}} \\  & \mathrm{=}\dfrac{{{\mathrm{a}}^{\mathrm{2}}}\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)}{{{\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}} \\  & \mathrm{=}\dfrac{\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)\left( {{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}} \right)}{{{\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)}^{\mathrm{2}}}} \\  & \mathrm{=}\dfrac{{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}}{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}} \\  \end{align} $ 

Hence, proved


5. Convert the following in the polar form:

$ \dfrac{\mathrm{1+7i}}{{{\mathrm{(2-i)}}^{\mathrm{2}}}} $                       $ \dfrac{\mathrm{1+3i}}{\mathrm{1-2i}} $  

Evaluate 

Ans:

Here,  $ \mathrm{z=}\dfrac{\mathrm{1+7i}}{{{\mathrm{(2-i)}}^{\mathrm{2}}}} $ 

 $ \begin{align} &\mathrm{=}\dfrac{\mathrm{1+7i}}{{{\mathrm{(2-i)}}^{\mathrm{2}}}}\mathrm{=}\dfrac{\mathrm{1+7i}}{\mathrm{4+}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{-4i}}\mathrm{=}\dfrac{\mathrm{1+7i}}{\mathrm{4-1-4i}} \\  & \mathrm{=}\dfrac{\mathrm{1+7i}}{\mathrm{3-4i}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{3+4i}}{\mathrm{3+4i}}\mathrm{=}\dfrac{\mathrm{3+4i+21i+28}{{\mathrm{i}}^{\mathrm{2}}}}{{{\mathrm{3}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{4}}^{\mathrm{2}}}} \\ &\mathrm{=}\dfrac{\mathrm{3+4i+21i-28}}{{{\mathrm{3}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{4}}^{\mathrm{2}}}}\mathrm{=}\dfrac{\mathrm{-25+25i}}{\mathrm{25}} \\  & \mathrm{=-1+i} \\ \end{align} $ 

Let  $ \mathrm{rcos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ rsin }\!\!\theta\!\!\text{ =1} $ 

Squaring and adding

$ \begin{align} & {{\mathrm{r}}^{\mathrm{2}}}\left( \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ } \right)\mathrm{=2} \\  & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{=2} \\  & \mathrm{r=}\sqrt{\mathrm{2}} \\  \end{align} $ 

 $ \begin{align} & \sqrt{\mathrm{2}}\mathrm{cos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{sin }\!\!\theta\!\!\text{ =1} \\  & \mathrm{cos }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{-1}}{\sqrt{\mathrm{2}}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ sin }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\  & \mathrm{ }\!\!\theta\!\!\text{ = }\!\!\pi\!\!\text{ -}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{=}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}} \\  & \mathrm{ }\!\!\!\!\text{ z=rcos }\!\!\theta\!\!\text{ +irsin }\!\!\theta\!\!\text{ } \\  \end{align} $ 

 $ \mathrm{=}\sqrt{\mathrm{2}}\mathrm{cos}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{+i}\sqrt{\mathrm{2}}\mathrm{sin}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{=}\sqrt{\mathrm{2}}\left( \mathrm{cos}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{+isin}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}} \right) $ 

This is the required polar form

Here,  $ \mathrm{z=}\dfrac{\mathrm{1+3i}}{\mathrm{1-2i}} $ 

$ \begin{matrix} \mathrm{=}\dfrac{\mathrm{1+3i}}{\mathrm{1-2i}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{1+2i}}{\mathrm{1+2i}}  \\ \mathrm{=}\dfrac{\mathrm{1+2i+3i-6}}{\mathrm{1+4}}  \\ \mathrm{=}\dfrac{\mathrm{-5+5i}}{\mathrm{5}}\mathrm{=-1+i}  \\ \end{matrix} $ 

Let  $ \mathrm{rcos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ rsin }\!\!\theta\!\!\text{ } $ 

 $ \begin{align} & \mathrm{=1 }\!\!~\!\!\text{ on squaring and adding, we obtain }\!\!~\!\!\text{ }{{\mathrm{r}}^{\mathrm{2}}}\left( \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ } \right) \\  & \mathrm{=1+1} \\  & {{\mathrm{r}}^{\mathrm{2}}}\left( \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ } \right)\mathrm{=2} \\  & {{\mathrm{r}}^{\mathrm{2}}}\mathrm{=2}\quad \left[ \mathrm{co}{{\mathrm{s}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ +si}{{\mathrm{n}}^{\mathrm{2}}}\mathrm{ }\!\!\theta\!\!\text{ =1} \right] \\  & \mathrm{r=}\sqrt{\mathrm{2}}\quad \mathrm{ }\!\![\!\!\text{  }\!\!~\!\!\text{ Conventionally, }\!\!~\!\!\text{ r0 }\!\!]\!\!\text{ } \\ \end{align} $ 

 $ \begin{align} & \mathrm{ }\!\!\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{cos }\!\!\theta\!\!\text{ =-1 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{sin }\!\!\theta\!\!\text{ =1} \\  & \mathrm{cos }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{-1}}{\sqrt{\mathrm{2}}}\mathrm{ }\!\!~\!\!\text{ and }\!\!~\!\!\text{ sin }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\  & \mathrm{ }\!\!\!\!\text{  }\!\!\theta\!\!\text{ = }\!\!\pi\!\!\text{ -}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{=}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}} \\  & \mathrm{ }\!\!\!\!\text{ z=rcos }\!\!\theta\!\!\text{ +irsin }\!\!\theta\!\!\text{ } \\  \end{align} $ 

 $ \mathrm{=}\sqrt{\mathrm{2}}\mathrm{cos}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{+i}\sqrt{\mathrm{2}}\mathrm{sin}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{=}\sqrt{\mathrm{2}}\left( \mathrm{cos}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{+isin}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}} \right) $ 

This is the required polar form


6. Solve the equation $ \mathrm{3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-4x+}\dfrac{\mathrm{20}}{\mathrm{3}}\mathrm{=0} $  

Evaluate 

Ans:

The given quadratic equation is  $ \mathrm{3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-4x+}\dfrac{\mathrm{20}}{\mathrm{3}}\mathrm{=0} $ 

On comparing this equation with  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

 $ \mathrm{a=9,b=-12 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=20} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=(-12}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ 9 }\!\!\times\!\!\text{ 20=144-720=-576} $ 

Therefore, the required solutions are

$ \begin{matrix} \dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-(12) }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-576}}}{\mathrm{2 }\!\!\times\!\!\text{ 9}}\mathrm{=}\dfrac{\mathrm{12 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{576}}\mathrm{i}}{\mathrm{18}}  \\ \mathrm{=}\dfrac{\mathrm{12 }\!\!\pm\!\!\text{ 24i}}{\mathrm{18}}\mathrm{=}\dfrac{\mathrm{6(2 }\!\!\pm\!\!\text{ 4i)}}{\mathrm{18}}\mathrm{=}\dfrac{\mathrm{2 }\!\!\pm\!\!\text{ 4i}}{\mathrm{3}}\mathrm{=}\dfrac{\mathrm{2}}{\mathrm{3}}\mathrm{ }\!\!\pm\!\!\text{ }\dfrac{\mathrm{4}}{\mathrm{3}}\mathrm{i}  \\ \end{matrix} $ 

Hence, solved

7. Solve the equation $ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-2x+}\dfrac{\mathrm{3}}{\mathrm{2}}\mathrm{=0} $  

 Evaluate 

Ans:

The given quadratic equation is  $ {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-2x+}\dfrac{\mathrm{3}}{\mathrm{2}}\mathrm{=0} $ 

On comparing this equation with  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

 $ \mathrm{a=2,b=-4 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=3} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=(-4}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 3=16-24=-8} $  

Therefore, the required solutions are

$ \begin{align} & \dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-(-4) }\!\!\pm\!\!\text{ }\sqrt{\mathrm{8}}}{\mathrm{2 }\!\!\times\!\!\text{ 2}}\mathrm{=}\dfrac{\mathrm{4 }\!\!\pm\!\!\text{ 2}\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{4}} \\ & \mathrm{=}\dfrac{\mathrm{2 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{2}}\mathrm{=1 }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i} \\  \end{align} $  

Hence, solved


8. Solve the equation

$ \mathrm{27}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-10x+1=0} $  

Evaluate 

Ans:

The given quadratic equation is  $ \mathrm{27}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-10x+1=0} $ 

On comparing this equation with  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

 $ \mathrm{a=27,b=-10 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=1} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=(-10}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ 27 }\!\!\times\!\!\text{ 1=100-108=-8} $  

Therefore, the required solutions are

$ \begin{align} & \dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-(-10) }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-8}}}{\mathrm{2 }\!\!\times\!\!\text{ 27}}\mathrm{=}\dfrac{\mathrm{10 }\!\!\pm\!\!\text{ 2}\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{54}} \\ & \mathrm{=}\dfrac{\mathrm{5 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{2}}\mathrm{i}}{\mathrm{27}}\mathrm{=}\dfrac{\mathrm{5}}{\mathrm{27}}\mathrm{ }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\mathrm{2}}}{\mathrm{27}}\mathrm{i} \\  \end{align} $  

Hence, solved


9. Solve the equation

$ \mathrm{21}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-28x+10=0} $  

Evaluate 

Ans:

The given quadratic equation is  $ \mathrm{21}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-28x+10=0} $ 

On comparing this equation with  $ \mathrm{a}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+bx+c=0} $ 

$ \mathrm{a=21,b=-28 }\!\!~\!\!\text{ and }\!\!~\!\!\text{ c=10} $ 

Therefore, the discriminant of the given equation is

$ \mathrm{D=}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-4ac=(-28}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{-4 }\!\!\times\!\!\text{ 21 }\!\!\times\!\!\text{ 10=-56} $  

Therefore, the required solutions are

$ \begin{align} & \dfrac{\mathrm{-b }\!\!\pm\!\!\text{ }\sqrt{\mathrm{D}}}{\mathrm{2a}}\mathrm{=}\dfrac{\mathrm{-(-28) }\!\!\pm\!\!\text{ }\sqrt{\mathrm{-56}}}{\mathrm{2 }\!\!\times\!\!\text{ 21}}\mathrm{=}\dfrac{\mathrm{28 }\!\!\pm\!\!\text{ }\sqrt{\mathrm{56}}\mathrm{i}}{\mathrm{42}} \\  & \mathrm{=}\dfrac{\mathrm{28 }\!\!\pm\!\!\text{ 2}\sqrt{\mathrm{14}}\mathrm{i}}{\mathrm{42}}\mathrm{=}\dfrac{\mathrm{2}}{\mathrm{3}}\mathrm{ }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\mathrm{14}}}{\mathrm{21}}\mathrm{i} \\  \end{align} $  

Hence, solved


10. If $ {{\mathrm{z}}_{\mathrm{1}}}\mathrm{=2-i,}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=1+i} $ Find $ \left| \dfrac{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{+}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}}{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{-}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}} \right| $ 

Evaluate 

Ans:

Complex numbers 

$ {{\mathrm{z}}_{\mathrm{1}}}\mathrm{=2-i,}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=1+i} $ 

$ \begin{matrix}  \mathrm{ }\!\!\!\!\text{ }\left| \dfrac{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{+}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}}{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{-}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}} \right|\mathrm{=}\left| \dfrac{\mathrm{(2-i)+(1+i)+1}}{\mathrm{(2-i)-(1+i)+1}} \right|  \\ \mathrm{=}\left| \dfrac{\mathrm{4}}{\mathrm{2-2i}} \right|\mathrm{=}\left| \dfrac{\mathrm{4}}{\mathrm{2(1-i)}} \right|  \\ \mathrm{=}\left| \dfrac{\mathrm{2}}{\mathrm{1-i}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{1+i}}{\mathrm{1+i}} \right|\mathrm{=}\left| \dfrac{\mathrm{2(1+i)}}{\left( {{\mathrm{1}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{i}}^{\mathrm{2}}} \right)} \right|  \\ \mathrm{=}\left| \dfrac{\mathrm{2(1+i)}}{\mathrm{1+1}} \right|\quad \left[ {{\mathrm{i}}^{\mathrm{2}}}\mathrm{=-1} \right]  \\ \mathrm{=}\left| \dfrac{\mathrm{2(1+i)}}{\mathrm{2}} \right|  \\ \end{matrix} $ 

$ \mathrm{= }\!\!|\!\!\text{ 1+i }\!\!|\!\!\text{ =}\sqrt{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}}}\mathrm{=}\sqrt{\mathrm{2}} $ 

Thus, the value of $ \left| \dfrac{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{+}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}}{{{\mathrm{z}}_{\mathrm{1}}}\mathrm{-}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{+1}} \right| $  is  $ \sqrt{\mathrm{2}} $


11. If  $ \mathrm{a+ib=}\dfrac{{{\mathrm{(x+i)}}^{\mathrm{2}}}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} $  

Prove that   $ {{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{=}\dfrac{{{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}}{{{\left( \mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}} $ 

Ans:

Expression 

$ \mathrm{a+ib=}\dfrac{{{\mathrm{(x+i)}}^{\mathrm{2}}}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} $ 

$ \begin{align} &\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{+2xi}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} \\ &\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-1+i2x}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} \\ &\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-1}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}}\mathrm{+i}\left( \dfrac{\mathrm{2x}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} \right) \\  \end{align} $ 

On comparing

$ \begin{align} & \mathrm{ }\!\!\!\!\text{ }{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{=}{{\left( \dfrac{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-1}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} \right)}^{\mathrm{2}}}\mathrm{+}{{\left( \dfrac{\mathrm{2x}}{\mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1}} \right)}^{\mathrm{2}}} \\ &\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{4}}}\mathrm{+1-2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+4}{{\mathrm{x}}^{\mathrm{2}}}}{{{\mathrm{(2x+1)}}^{\mathrm{2}}}} \\ &\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{4}}}\mathrm{+1+2}{{\mathrm{x}}^{\mathrm{2}}}}{{{\left( \mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}} \\  & \mathrm{=}\dfrac{{{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}}{{{\left( \mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}} \\  & \mathrm{ }\!\!\!\!\text{ }{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{=}\dfrac{{{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}}{{{\left( \mathrm{2}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+1} \right)}^{\mathrm{2}}}} \\  \end{align} $ 

Hence, proved


12. Let $ {{\mathrm{z}}_{\mathrm{1}}}\mathrm{=2-i,}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=-2+i} $ 

Find    $ \begin{align} & \mathrm{Re}\left( \dfrac{{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}}{{{{\mathrm{\bar{z}}}}_{\mathrm{1}}}} \right) \\  & \mathrm{Im}\left( \dfrac{\mathrm{1}}{{{\mathrm{z}}_{\mathrm{1}}}{{{\mathrm{\bar{z}}}}_{\mathrm{1}}}} \right) \\  \end{align} $ 

Ans:

Complex numbers $ \begin{align} &{{\mathrm{z}}_{\mathrm{1}}}\mathrm{=2-i,}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=-2+i} \\ &{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}\mathrm{=(2-i)(-2+i)=-4+2i+2i-}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{=-4+4i-(-1)=-3+4i} \\ & \overline{{{\mathrm{z}}_{\mathrm{1}}}}\mathrm{=2+i} \\  & \mathrm{ }\!\!\!\!\text{ }\dfrac{{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}}{\overline{{{\mathrm{z}}_{\mathrm{1}}}}}\mathrm{=}\dfrac{\mathrm{-3+4i}}{\mathrm{2+i}} \\  \end{align} $ 

On multiplying numerator and denominator by  $ \left( 2-i \right) $ , we obtain 

$ \begin{align} &\dfrac{{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}}{\overline{{{\mathrm{z}}_{\mathrm{1}}}}}\mathrm{=}\dfrac{\mathrm{(-3+4i)(2-i)}}{\mathrm{(2+i)(2-i)}}\mathrm{=}\dfrac{\mathrm{-6+3i+8i-4}{{\mathrm{i}}^{\mathrm{2}}}}{{{\mathrm{2}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}}}\mathrm{=}\dfrac{\mathrm{-6+11i-4(-1)}}{{{\mathrm{2}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}}} \\ &\mathrm{=}\dfrac{\mathrm{-2+11i}}{\mathrm{5}}\mathrm{=}\dfrac{\mathrm{-2}}{\mathrm{5}}\mathrm{+}\dfrac{\mathrm{11}}{\mathrm{5}}\mathrm{i} \\  \end{align} $ 

On comparing real parts, we obtain

$ \begin{align} & \mathrm{Re}\left( \dfrac{{{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}}}{{{{\mathrm{\bar{z}}}}_{\mathrm{1}}}} \right)\mathrm{=}\dfrac{\mathrm{-2}}{\mathrm{5}} \\ & \mathrm{ }\!\!~\!\!\text{ }\dfrac{\mathrm{1}}{{{\mathrm{z}}_{\mathrm{1}}}{{{\mathrm{\bar{z}}}}_{\mathrm{1}}}}\mathrm{=}\dfrac{\mathrm{1}}{\mathrm{(2-i)(2+i)}}\mathrm{=}\dfrac{\mathrm{1}}{{{\mathrm{(2)}}^{\mathrm{2}}}\mathrm{+(1}{{\mathrm{)}}^{\mathrm{2}}}}\mathrm{=}\dfrac{\mathrm{1}}{\mathrm{5}} \\  \end{align} $ 

On comparing imaginary parts, we obtain

$ \mathrm{Im}\left( \dfrac{\mathrm{1}}{{{\mathrm{z}}_{\mathrm{1}}}{{{\mathrm{\bar{z}}}}_{\mathrm{1}}}} \right)\mathrm{=0} $ 

Hence, solved


13. Find the modulus and argument of the complex number 

$ \dfrac{\mathrm{1+2i}}{\mathrm{1-3i}} $ 

Evaluate 

Ans:

Let  $ \mathrm{z=}\dfrac{\mathrm{1+2i}}{\mathrm{1-3i}} $  , then 

$ \begin{align} & \mathrm{z=}\dfrac{\mathrm{1+2i}}{\mathrm{1-3i}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{1+3i}}{\mathrm{1+3i}}\mathrm{=}\dfrac{\mathrm{1+3i+2i+6}{{\mathrm{i}}^{\mathrm{2}}}}{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{3}}^{\mathrm{2}}}}\mathrm{=}\dfrac{\mathrm{1+5i+6(-1)}}{\mathrm{1+9}} \\ &\mathrm{=}\dfrac{\mathrm{-5+5i}}{\mathrm{10}}\mathrm{=}\dfrac{\mathrm{-5}}{\mathrm{10}}\mathrm{+}\dfrac{\mathrm{5i}}{\mathrm{10}}\mathrm{=}\dfrac{\mathrm{-1}}{\mathrm{2}}\mathrm{+}\dfrac{\mathrm{1}}{\mathrm{2}}\mathrm{i} \\  \end{align} $ 

Let  $ \mathrm{z=rcos }\!\!\theta\!\!\text{ +irsin }\!\!\theta\!\!\text{ } $ 

i.e.,  $ \mathrm{rcos }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{-1}}{\mathrm{2}} $  and  $ \mathrm{rsin }\!\!\theta\!\!\text{ =}\dfrac{\mathrm{1}}{\mathrm{2}} $ 

On squaring and adding, we obtain

$ \mathrm{r=}\dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}} $ 

$ \begin{align} & \mathrm{ }\!\!~\!\!\text{  }\!\!\!\!\text{  }\!\!\theta\!\!\text{ = }\!\!\pi\!\!\text{ -}\dfrac{\mathrm{ }\!\!\pi\!\!\text{ }}{\mathrm{4}}\mathrm{=}\dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}} \\ & \mathrm{ }\!\![\!\!\text{ As }\!\!~\!\!\text{  }\!\!\theta\!\!\text{  }\!\!~\!\!\text{ lies in the II quadrant }\!\!]\!\!\text{ } \\  \end{align} $ 

Therefore, the modulus and argument of the given complex number are  $ \dfrac{\mathrm{1}}{\sqrt{\mathrm{2}}} $  and  $ \dfrac{\mathrm{3 }\!\!\pi\!\!\text{ }}{\mathrm{4}} $ respectively


14. Find the real numbers  $ \mathrm{x }\!\!\And\!\!\text{ y} $  if  $ \left( \mathrm{x-iy} \right)\left( \mathrm{3+5i} \right) $  is the conjugate of  $ \mathrm{-6-24i} $ 

Ans:

Let  $ \mathrm{z=}\left( \mathrm{x-iy} \right)\left( \mathrm{3+5i} \right) $ 

$ \begin{align} &\mathrm{z=3x+5xi-3yi-5y}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{=3x+5xi-3yi+5y=(3x+5y)+i(5x-3y)} \\  & \mathrm{ }\!\!\!\!\text{ \bar{z}=(3x+5y)-i(5x-3y)} \\  \end{align} $ 

It is given that,  $ \overline{\mathrm{z}}\mathrm{=-6-24i} $ 

$ \mathrm{ }\!\!\!\!\text{ (3x+5y)-i(5x-3y)=-6-24i} $ 

Equating real and imaginary parts, we obtain

$ \begin{matrix} \mathrm{3x+5y=-6}\quad \mathrm{}..\mathrm{(i)}  \\ \mathrm{5x-3y=24}...\mathrm{(ii)}  \\ \end{matrix} $ 

On solving we will get 

$ \begin{align} & \mathrm{3(3)+5y=-6} \\ & \mathrm{5y=-6-9=-15} \\  & \mathrm{y=-3} \\  \end{align} $ 

Thus, the values of  $ \mathrm{x and y are 3 and -3} $ respectively


15. Find the modulus of  $ \dfrac{\mathrm{1+i}}{\mathrm{1-i}}\mathrm{-}\dfrac{\mathrm{1-i}}{\mathrm{1+i}} $ 

Evaluate  

Ans:

Expression 

$ \begin{align} &\dfrac{\mathrm{1+i}}{\mathrm{1-i}}\mathrm{-}\dfrac{\mathrm{1-i}}{\mathrm{1+i}}\mathrm{=}\dfrac{{{\mathrm{(1+i)}}^{\mathrm{2}}}\mathrm{-(1-i}{{\mathrm{)}}^{\mathrm{2}}}}{\mathrm{(1-i)(1+i)}} \\ &\mathrm{=}\dfrac{\mathrm{1+}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{+2i-1-}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{+2i}}{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}}} \\ &\mathrm{=}\dfrac{\mathrm{4i}}{\mathrm{2}}\mathrm{=2i} \\  & \left| \dfrac{\mathrm{1+i}}{\mathrm{1-i}}\mathrm{-}\dfrac{\mathrm{1-i}}{\mathrm{1+i}} \right|\mathrm{= }\!\!|\!\!\text{ 2i }\!\!|\!\!\text{ =}\sqrt{{{\mathrm{2}}^{\mathrm{2}}}}\mathrm{=2} \\  \end{align} $  

Here we get the answer

16. Find the modulus of  $ {{\mathrm{(x+iy)}}^{\mathrm{3}}}\mathrm{=u+iv} $  

Than show that    $ \dfrac{\mathrm{u}}{\mathrm{x}}\mathrm{+}\dfrac{\mathrm{v}}{\mathrm{y}}\mathrm{=4}\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \right) $ 

Ans:

$ \begin{align} & {{\mathrm{(x+iy)}}^{\mathrm{3}}}\mathrm{=u+iv} \\ &\mathrm{}{{\mathrm{x}}^{\mathrm{3}}}\mathrm{+(iy}{{\mathrm{)}}^{\mathrm{3}}}\mathrm{+3 }\!\!\times\!\!\text{ x }\!\!\times\!\!\text{ iy(x+iy)=u+iv} \\ &\mathrm{}{{\mathrm{x}}^{\mathrm{3}}}\mathrm{+}{{\mathrm{i}}^{\mathrm{3}}}{{\mathrm{y}}^{\mathrm{3}}}\mathrm{+3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{yi+3x}{{\mathrm{y}}^{\mathrm{2}}}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{=u+iv} \\ &\mathrm{}{{\mathrm{x}}^{\mathrm{3}}}\mathrm{-i}{{\mathrm{y}}^{\mathrm{3}}}\mathrm{+3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{yi-3x}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{=u+iv} \\ & \mathrm{}\left( {{\mathrm{x}}^{\mathrm{3}}}\mathrm{-3x}{{\mathrm{y}}^{\mathrm{2}}} \right)\mathrm{+i}\left( \mathrm{3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{y-}{{\mathrm{y}}^{\mathrm{3}}} \right)\mathrm{=u+iv} \\  \end{align} $ 

On equating real and imaginary

$ \begin{align} &\mathrm{u=}{{\mathrm{x}}^{\mathrm{3}}}\mathrm{-3x}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{,v=3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{y-}{{\mathrm{y}}^{\mathrm{3}}} \\ &\dfrac{\mathrm{u}}{\mathrm{x}}\mathrm{+}\dfrac{\mathrm{v}}{\mathrm{y}}\mathrm{=}\dfrac{{{\mathrm{x}}^{\mathrm{3}}}\mathrm{-3x}{{\mathrm{y}}^{\mathrm{2}}}}{\mathrm{x}}\mathrm{+}\dfrac{\mathrm{3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{y-}{{\mathrm{y}}^{\mathrm{3}}}}{\mathrm{y}} \\  & \mathrm{=}\dfrac{\mathrm{x}\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-3}{{\mathrm{y}}^{\mathrm{2}}} \right)}{\mathrm{x}}\mathrm{+}\dfrac{\mathrm{y}\left( \mathrm{3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \right)}{\mathrm{y}} \\ &\mathrm{=}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-3}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{+3}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \\ &\mathrm{=4}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{-4}{{\mathrm{y}}^{\mathrm{2}}} \\  & \mathrm{=4}\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \right) \\ &\dfrac{\mathrm{u}}{\mathrm{x}}\mathrm{+}\dfrac{\mathrm{v}}{\mathrm{y}}\mathrm{=4}\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{-}{{\mathrm{y}}^{\mathrm{2}}} \right) \\ \end{align} $ 

Hence, proved

 

17. If  $ \mathrm{ }\!\!\alpha\!\!\text{ and }\!\!\beta\!\!\text{ } $  are different complex numbers with  $ \left| \mathrm{ }\!\!\beta\!\!\text{ } \right|\mathrm{=1} $  , then find  $ \left| \dfrac{\mathrm{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ }}{\mathrm{1-}\overline{\mathrm{ }\!\!\alpha\!\!\text{ }}\mathrm{ }\!\!\beta\!\!\text{ }} \right|\mathrm{=1} $ 

Ans:

Let  $ \mathrm{ }\!\!\alpha\!\!\text{ =a+ib }\!\!\And\!\!\text{  }\!\!\beta\!\!\text{ =x+iy} $ 

It is given that,  $ \left| \mathrm{ }\!\!\beta\!\!\text{ } \right|\mathrm{=1} $ 

$ \begin{align} &\sqrt{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}}}\mathrm{=1} \\ &\mathrm{}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{=1}..\left| \dfrac{\mathrm{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ }}{\mathrm{1-\bar{ }\!\!\alpha\!\!\text{ }}} \right|\mathrm{=}\left| \dfrac{\mathrm{(x+iy)-(a+ib)}}{\mathrm{1-(a-ib)(x+iy)}} \right| \\  & \mathrm{=}\left| \dfrac{\mathrm{(x-a)+i(y-b)}}{\mathrm{1-(ax+aiy-ibx+by)}} \right| \\   & \mathrm{=}\left| \dfrac{\mathrm{(x-a)+i(y-b)}}{\mathrm{(1-ax-by)+i(bx-ay)}} \right| \\  & \mathrm{=}\left| \dfrac{\mathrm{(x-a)+i(y-b)}}{\mathrm{(1-ax-by)+i(bx-ay)}} \right| \\  \end{align} $ 

$ \begin{align} & \mathrm{=}\dfrac{\sqrt{{{\mathrm{(x-a)}}^{\mathrm{2}}}\mathrm{+(y-b}{{\mathrm{)}}^{\mathrm{2}}}}}{\sqrt{{{\mathrm{(1-ax-by)}}^{\mathrm{2}}}\mathrm{+(bx-ay}{{\mathrm{)}}^{\mathrm{2}}}}} \\ &\mathrm{=}\dfrac{\sqrt{{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{a}}^{\mathrm{2}}}\mathrm{-2ax+}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-2by}}}{\sqrt{\mathrm{1+}{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{-2ax+2abxy-2by+}{{\mathrm{b}}^{\mathrm{2}}}{{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{a}}^{\mathrm{2}}}{{\mathrm{y}}^{\mathrm{2}}}\mathrm{-2abxy}}} \\  & \mathrm{=}\dfrac{\sqrt{\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}} \right)\mathrm{+}{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\mathrm{-2ax-2by}}}{\sqrt{\mathrm{1+}{{\mathrm{a}}^{\mathrm{2}}}\left( {{\mathrm{x}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{y}}^{\mathrm{2}}} \right)\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}\left( {{\mathrm{y}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{x}}^{\mathrm{2}}} \right)\mathrm{-2ax-2by}}} \\  \end{align} $ 

$ \left| \dfrac{\mathrm{ }\!\!\beta\!\!\text{ - }\!\!\alpha\!\!\text{ }}{\mathrm{1-}\overline{\mathrm{ }\!\!\alpha\!\!\text{ }}\mathrm{ }\!\!\beta\!\!\text{ }} \right|\mathrm{=1} $


18. Find the number of non-zero integral solutions of the equation  $ {{\left| \mathrm{1-i} \right|}^{\mathrm{x}}}\mathrm{=}{{\mathrm{2}}^{\mathrm{x}}} $ 

Ans:

Equation 

$ \begin{align} & \mathrm{ }\!\!|\!\!\text{ 1-i}{{\mathrm{ }\!\!|\!\!\text{ }}^{\mathrm{x}}}\mathrm{=}{{\mathrm{2}}^{\mathrm{x}}} \\  & {{\left( \sqrt{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+(-1}{{\mathrm{)}}^{\mathrm{2}}}} \right)}^{\mathrm{x}}}\mathrm{=}{{\mathrm{2}}^{\mathrm{x}}} \\  & {{\mathrm{(}\sqrt{\mathrm{2}}\mathrm{)}}^{\mathrm{x}}}\mathrm{=}{{\mathrm{2}}^{\mathrm{x}}} \\  & {{\mathrm{2}}^{\mathrm{x/2}}}\mathrm{=}{{\mathrm{2}}^{\mathrm{x}}} \\  & \dfrac{\mathrm{x}}{\mathrm{2}}\mathrm{=x} \\  & \mathrm{x=2x} \\  & \mathrm{x=0} \\  \end{align} $ 

Thus,  $ \mathrm{0} $ is the only integral solution of the given equation. Therefore, the number of nonzero integral solutions of the given equation is  $ \mathrm{0} $.


19. If  $ \mathrm{(a+ib)(c+id)(e+if)(g+ih)=A+iB} $ Then show that $ \left( {{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}} \right)\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)\left( {{\mathrm{e}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{f}}^{\mathrm{2}}} \right)\left( {{\mathrm{g}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{h}}^{\mathrm{2}}} \right)\mathrm{=}{{\mathrm{A}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{B}}^{\mathrm{2}}} $  

Ans:

Expression 

$ \begin{align} & \mathrm{(a+ib)(c+id)(e+if)(g+ih)=A+iB} \\  & \mathrm{ }\!\!\!\!\text{  }\!\!|\!\!\text{ (a+ib)(c+id)(e+if)(g+ih) }\!\!|\!\!\text{ = }\!\!|\!\!\text{ A+iB }\!\!|\!\!\text{ } \\  & \mathrm{ }\!\!|\!\!\text{ (a+ib) }\!\!|\!\!\text{  }\!\!\times\!\!\text{  }\!\!|\!\!\text{ (c+id) }\!\!|\!\!\text{  }\!\!\times\!\!\text{  }\!\!|\!\!\text{ (e+if) }\!\!|\!\!\text{  }\!\!\times\!\!\text{  }\!\!|\!\!\text{ (g+ih) }\!\!|\!\!\text{ = }\!\!|\!\!\text{ A+iB }\!\!|\!\!\text{ }\quad \mathrm{Q}\left[ \left| {{\mathrm{z}}_{\mathrm{1}}}{{\mathrm{z}}_{\mathrm{2}}} \right|\mathrm{=}\left| {{\mathrm{z}}_{\mathrm{1}}} \right|\left| {{\mathrm{z}}_{\mathrm{2}}} \right| \right] \\ &\sqrt{{{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}}}\mathrm{ }\!\!\times\!\!\text{ }\sqrt{{{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}}}\mathrm{ }\!\!\times\!\!\text{ }\sqrt{{{\mathrm{e}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{f}}^{\mathrm{2}}}}\mathrm{ }\!\!\times\!\!\text{ }\sqrt{{{\mathrm{g}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{h}}^{\mathrm{2}}}}\mathrm{=}\sqrt{{{\mathrm{A}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{B}}^{\mathrm{2}}}} \\  \end{align} $ 

By squaring 

$ \left( {{\mathrm{a}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{b}}^{\mathrm{2}}} \right)\left( {{\mathrm{c}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{d}}^{\mathrm{2}}} \right)\left( {{\mathrm{e}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{f}}^{\mathrm{2}}} \right)\left( {{\mathrm{g}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{h}}^{\mathrm{2}}} \right)\mathrm{=}{{\mathrm{A}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{B}}^{\mathrm{2}}} $ 

Hence, proved


20. If  $ {{\left( \dfrac{\mathrm{1+i}}{\mathrm{1-i}} \right)}^{\mathrm{m}}}\mathrm{=1} $  

Then find the least positive integral value of  $ m $ 

Ans

$ \begin{align} & {{\left( \dfrac{\mathrm{1+i}}{\mathrm{1-i}} \right)}^{\mathrm{m}}}\mathrm{=1} \\  & {{\left( \dfrac{\mathrm{1+i}}{\mathrm{1-i}}\mathrm{ }\!\!\times\!\!\text{ }\dfrac{\mathrm{1+i}}{\mathrm{1+i}} \right)}^{\mathrm{m}}}\mathrm{=1} \\  & {{\left( \dfrac{{{\mathrm{(1+i)}}^{\mathrm{2}}}}{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{1}}^{\mathrm{2}}}} \right)}^{\mathrm{m}}}\mathrm{=1} \\  & {{\left( \dfrac{{{\mathrm{1}}^{\mathrm{2}}}\mathrm{+}{{\mathrm{i}}^{\mathrm{2}}}\mathrm{+2i}}{\mathrm{2}} \right)}^{\mathrm{m}}}\mathrm{=1} \\ \end{align} $ 

$ \begin{align} & {{\left( \dfrac{\mathrm{1-1+2i}}{\mathrm{2}} \right)}^{\mathrm{m}}}\mathrm{=1} \\  & {{\left( \dfrac{\mathrm{2i}}{\mathrm{2}} \right)}^{\mathrm{m}}}\mathrm{=1} \\  & {{\mathrm{i}}^{\mathrm{m}}}\mathrm{=1} \\  & {{\mathrm{i}}^{\mathrm{m}}}\mathrm{=}{{\mathrm{i}}^{\mathrm{4k}}} \\ \end{align} $  

$ \mathrm{m=4k} $  , where  $ \mathrm{k} $  is some integer

Therefore, the least positive is one

Thus, the least positive integral value of  $ \mathrm{m} $  is  $ \mathrm{4=}\left( \mathrm{4 }\!\!\times\!\!\text{ 1} \right) $


NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

NCERT Solutions For Class 11 Math Chapter 5 Complex Numbers and Quadratic Equations are available here in the pdf format. These NCERT Solutions of Class 11 Math Chapter 5 help the students in solving the problems quickly, accurately and efficiently. Also, our subject expert teachers have provided step by step solutions for all NCERT problems, thereby ensuring students understand them and clear their academic examinations with good grades. These NCERT Solutions are provided by Vedantu with an aim to help students who aim to clear their examination  even with the last-minute preparations. However, NCERT Solutions For Class 11 Math Chapter 5 are focused on mastering the concepts along with gaining broader knowledge.


Topics Covered In Class 11 Math Chapter 5

5.1: Introduction

5.2: Complex Number

5.3: Algebra of Complex Number: Addition, Subtraction, Multiplication, & Division of Two Complex Numbers

  • Power of i

  • The Square Root of Negative Real Numbers

  • Identities

5.4: Modulus and Conjugate of Complex Number

5.5: Argand Plane And Polar Representation

  • Polar Representation of Complex Number

5.6: Quadratic Equation


What Are Complex Numbers?

A number in the form of a + ib, where a and b are real numbers, are termed as complex numbers. Here, ‘a’ is the real part and ‘b’ is the imaginary part of the complex number.


What Are Quadratic Equations?

Quadratic equation is an equation of degree 2. It implies that the highest component of this function is 2. The standard form of quadratic equation is given as: 


ax2 + bx + c = 0


Here, a,b,and c R, a 0, b2 - 4ac < 0 are given by -b 4ac - b2 i2a


Some Examples of Complex Number Class 11 Solutions: Question 1 to 10

The solutions provide a step-by-step explanation in which the given complex numbers like (5i) (3/5i) and i9 +i19 are solved and expressed in terms of a+ib. The solved sums covered in these NCERT Solutions provide a detailed stepwise explanation so that students can easily grasp the mathematical formulas and the concept of these sums.

Complex Numbers Class 11 Solutions: Questions 11 to 13

These solutions provide a detailed description of the equations with which the multiplicative inverse of the given numbers 4-3i, Ö5+3i, and -i are extracted. The step-by-step explanations help a student to grasp the details of the chapter better.

NCERT Solutions for Class 11 Maths Chapter 5 NCERT Solutions of Exercise 5.2: Question 1 

The Complex Number NCERT Solutions help students to understand the equations and formulas the are required to find the modulus and argument of the complex number Z= -1-i

Ö3 respectively. By learning the steps, the students will be able to solve any related sums easily and improve their problem-solving skills.

Marks Weightage of Complex No. Class 11

Ch 5 Maths Class 11 complex numbers form a part of Unit 2- Algebra of CBSE Class 11 Maths syllabus, which has a weightage of 30 marks (out of total 80 marks). Therefore, students should keenly follow NCERT Solutions for Class 11 Maths Chapter 5, so that they can have a better conceptual understanding of the Algebraic equations and solutions to answer a wide category of exercises by applying the mathematical formulas.

Benefits of Complex Numbers Class 11 NCERT PDF

Complex Numbers Class 11 solutions NCERT PDF are beneficial in several ways.

  • Solved and explained by expert mathematicians

  • Updated to latest CBSE syllabus

  • Step by step solutions

  • Detailed equations and theorems

  • Helpful for self-study and doubt clearance

We cover all exercises in the chapter given below:- 

Therefore, every CBSE student must solve and practice NCERT Solutions for Class 11 Maths Chapter 5 so that they can get a comprehensive understanding of the chapter and apply the formulas and theorems effectively to solve the questions asked in the exam.


Benefits of Referring to NCERT Solutions Provided by Vedantu

We at Vedantu provide you with all the study material that you are searching for in your preparation for the exams. Here you will get NCERT Solutions, syllabus, previous year's paper’s solutions for exams, important questions etc.

The syllabus is provided here as per the CBSE guidelines. Solutions of NCERT are prepared by highly-experienced teachers.  The benefits of Vedantu are as follows.

  • The Solution in Easy Language

The NCERT Solutions of each chapter are prepared by different experts and scholars in the subject matter. The study materials offered by Vedantu are made available to students after rigorous research to ensure that all the given inputs are authentic and to the point.

  • Focus on Fundamental Concepts

NCERT Class 11 chapter-wise solutions not only cover all the topics in the syllabus but also vividly describe all the fundamental and basic concepts required to understand these topics.

  • Sufficient Material to Practice

Preparation for any exam is incomplete without practice. Students are required to practice questions in order to perform well in examinations. If you need any other study material, you can visit our website.

  • Important Topics

Important topics given in the chapter are discussed from the point of view of examination. You can get important topics in each chapter in Vedantu.

  • Better Preparation

Class 11 Maths NCERT Solutions will resolve the doubts of students quickly and their preparation for examinations will be boosted. With the help of these NCERT Solutions, students will grab complex concepts quickly.


NCERT Solutions for Class 11 Maths Chapters

 

Along with this, students can also download additional study materials provided by Vedantu, for Chapter 5 of CBSE Class 11 Maths Solutions –

Conclusion 

The NCERT Solutions for Class 11 Maths Chapter 5 is important for students to improve their performance. The important topics covered in the chapter are given above. The solutions are prepared by our experts in easy language so that students can easily understand them. It will surely help students. If you need any further assistance or required study material on other subjects, visit our website. 

FAQs on NCERT Solutions for Class 11 Maths Chapter 5 - Complex Numbers And Quadratic Equations

1. What are the exercises and topics covered under NCERT Complex Number Class 11 Maths Chapter 5?

The Ch 5 Maths Class 11 NCERT Solutions consist of solved exercises that cover critical equations related to complex numbers and quadratic equations. These NCERT Solutions provide clarity on the theorems and concepts of Complex Numbers. There are three sets of exercises in this chapter for which the solutions are given in this PDF. Exercise 5.1 has 14 questions with answers and covers topics like finding the multiplicative inverse or expressing a set of numbers into complex numbers.


Next is Exercise 5.2 that comprises 8 questions and covers topics like finding the modulus and argument of a given set of numbers. Lastly, Exercise 5.3 comprises 10 questions and provides step by step solutions for solving various quadratic equations. 

2. What is the marks distribution for Class 11 Maths?

The marking distribution for Class 11 Maths is discussed here. There are six units in which the mark weightage is distributed. First is Sets and Functions which have a marks weightage of 60 marks, the second unit is Algebra and has a marks weightage of 30.


Unit three covers the topic of Coordinate geometry and has a marks allocation of 10. Unit 4 covers Calculus and has a marks weightage of 30.


This is followed by unit 5 and unit 6 comprising mathematical reasoning, statistics, and probability with a mark’s distribution of 10 and 30 respectively. The total period under which all these chapters are divided is 240. 

3. On which website can we get the most reliable Complex Number Class 11 NCERT Solutions?

The most reliable and accurate NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are available on Vedantu. Students do not need to pay any additional charges for downloading the solutions if they already have a registration. These solutions are prepared and compiled by subject-matter experts who have considerable years of experience in handling the CBSE syllabus and therefore offer the most advanced and detailed solutions to the exercises. 

4. How can I ace Chapter 5 of Class 11 Maths?

You can easily ace Chapter - Complex Numbers and Quadratic Equations of Class 11 Maths. What you need is a strategy that you must follow consistently to get good grades. The best strategy is using NCERT Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations available on Vedantu. The NCERT Solutions PDF is very comprehensive and covers all the questions in detail. Practice all the exercises from the PDF to get full marks.

5. What are the important sub-topics that can come for exams from Class 11 Maths Chapter 5?

The important topics that need to be covered in NCERT Class 11 Maths Chapter 5 are -  Complex Numbers, 

  • Addition of Two Complex Numbers, 

  • Difference Between Two Complex Numbers, 

  • Multiplication and Division of Two Complex Numbers, 

  • Power of i (iota), 

  • Identities, 

  • Modulus and Conjugate of Complex Numbers, 

  • Argand Plane 

Solve all the examples and exercise questions thoroughly to complete your preparation for your test.

6. What are some of the properties of the multiplication of complex numbers?

Some multiplication properties of complex numbers are closure law (the product of two complex numbers is also a complex number), commutative law (product of x1 and x2= product of x2 and x1), associative law (for complex numbers x1, x2,x3, (x1 x2) x3 = x1 (x2x3)), multiplicative identity (for every complex number x, x multiplied by 1 is 1), multiplicative identity (x.1/x is 1), and distributive law (x1(x2,x3)=  x1x2, x1x3).

7. How should I prepare for NCERT Class 11 Maths Chapter 5?

You can download the NCERT Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations from Vedantu and practice each and every example and solved exercises including the miscellaneous questions. The solutions to all questions are given step by step so that you understand everything very clearly. All the answers have been provided by experts who have curated precise content for you to practice. These solutions are available at free of cost on Vedantu’s website(vedantu.com) and mobile app.

8. What is an argand plane?

An argand plane or a complex plane is a graphical representation of complex numbers that are plotted along x-axis and y-axis. The x-axis in an argand plane is understood as the real axis and the y-axis in an argand plane is called an imaginary axis. To get answers to more such questions from this chapter, you can get access to the NCERT Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations available only on Vedantu.