Important Questions for CBSE Class 11 Maths Chapter 5 - Complex Numbers and Quadratic Equations

Very Short Answer Questions: (1 Marks)

1. Evaluate the value of ${{i}^{-39}}$.

Ans:

Let us solve the given expression further –

${{i}^{-39}}={{i}^{-38-1}}$

$\Rightarrow {{i}^{-39}}={{({{i}^{2}})}^{-19}}.\dfrac{1}{i}$

$\Rightarrow {{i}^{-39}}={{(-1)}^{-19}}.\dfrac{1}{i}$

$\Rightarrow {{i}^{-39}}=-\dfrac{1}{i}\times \dfrac{i}{i}$

Therefore,

${{i}^{-39}}=i$

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

Ans:

Let us further solve the given quadratic equation ${{x}^{2}}+x+\dfrac{1}{\sqrt{2}}=0$.

$\Rightarrow \sqrt{2}{{x}^{2}}+\sqrt{2}x+1=0$.

Hence, we have $a=\sqrt{2}$,

$b=\sqrt{2}$, and

$c=1$.

We will find the roots by the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Therefore,

\[\Rightarrow x=\dfrac{-\sqrt{2}\pm \sqrt{{{(\sqrt{2})}^{2}}-4\sqrt{2}}}{2\sqrt{2}}\]

\[\Rightarrow x=\dfrac{-\sqrt{2}\pm \sqrt{2-4\sqrt{2}}}{2\sqrt{2}}\]

\[\Rightarrow x=\dfrac{-\sqrt{2}\pm \sqrt{2}\sqrt{1-2\sqrt{2}}}{2\sqrt{2}}\]

\[\Rightarrow x=\dfrac{-1\pm i\sqrt{2\sqrt{2}-1}}{2}\]

3. If ${{\left( \dfrac{1+i}{1-i} \right)}^{m}}=1$, then find the least positive integral value of $m$.

Ans:

Let us simplify the given expression –

${{\left( \dfrac{1+i}{1-i} \right)}^{m}}={{\left( \dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i} \right)}^{m}}$

$\Rightarrow {{\left( \dfrac{1+2i-1}{1+1} \right)}^{m}}$

$\Rightarrow {{\left( \dfrac{2i}{2} \right)}^{m}}$

$\Rightarrow {{i}^{m}}=1$

Now, we know that ${{i}^{4n}}=1$.

Therefore, ${{i}^{m}}={{i}^{4n}}$

$\Rightarrow m=4n$.

Where $n$ is an integer.

Hence, the least positive integral value of $m$ is $4$.

4. Evaluate the expression ${{(1+i)}^{4}}$.

Ans:

Let us simplify the given expression –

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

\[\Rightarrow {{\left( 1+i2+{{i}^{2}} \right)}^{2}}\]

\[\Rightarrow {{\left( 1+i2-1 \right)}^{2}}\]

\[\Rightarrow 4{{i}^{2}}\]

\[\Rightarrow -4\]

Hence, \[{{(1+i)}^{4}}=-4+0i\].

5. Find the modulus of \[\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}\].

Ans:

Let us simplify the given expression –

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

\[= \dfrac{\left( 1+2i-1 \right)-\left( 1-2i-1 \right)}{1+1}\]

\[= \dfrac{2i+2i}{2}\]

\[\Rightarrow z=2i\]

Hence, the modulus will be –

\[\left| z \right|=\sqrt{{{0}^{2}}+{{2}^{2}}}\]

\[\Rightarrow \left| z \right|=2\].

6. Express in the form of a+ib: \[{{\left( 1+3i \right)}^{-1}}\].

Ans:

Let us simplify the given expression –

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

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

\[= \dfrac{1-3i}{1-9{{i}^{2}}}\]

\[= \dfrac{1-3i}{10}\]

Therefore, \[{{\left( 1+3i \right)}^{-1}}=\dfrac{1}{10}-i\dfrac{3}{10}\].

7. Explain the fallacy in $-1=i.i=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt{1}=1$.

Ans:

As, the given equation shows that \[\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}\] is a fallacy because \[\sqrt{-1}\sqrt{-1}={{(\sqrt{-1})}^{2}}\]

\[\Rightarrow \sqrt{-1}\sqrt{-1}=-1\] is correct.

8. Find the conjugate of $\dfrac{1}{2-3i}$.

Ans:

Let us further solve the expression as –

\[\dfrac{1}{2-3i}=\dfrac{1}{2-3i}\times \dfrac{2+3i}{2+3i}\]

\[= \dfrac{2+3i}{4-9{{i}^{2}}}\]

\[= \dfrac{2+3i}{4+9}\]

\[= \dfrac{2+3i}{13}\]

Therefore, the conjugate will be \[\dfrac{2-3i}{13}\].

9. Find the conjugate of $-3i-5$.

Ans:

Let us consider a complex number \[z=-3i-5\].

Therefore, the conjugate will be \[\bar{z}=3i-5\].

10. Let ${{z}_{1}}=2-i$ and

${{z}_{2}}=-2+i$, then find \[\operatorname{Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{{{{\bar{z}}}_{1}}} \right)\].

Ans:

Given we have ${{z}_{1}}=2-i$ and

${{z}_{2}}=-2+i$.

Therefore, ${{z}_{1}}{{z}_{2}}=\left( 2-i \right)\left( -2+i \right)$

\[\Rightarrow {{z}_{1}}{{z}_{2}}=-4+2i+2i+1\]

\[\Rightarrow {{z}_{1}}{{z}_{2}}=-3+4i\].

Also, ${{\bar{z}}_{1}}=2+i$.

Hence, $\dfrac{{{z}_{1}}{{z}_{2}}}{{{{\bar{z}}}_{1}}}=\dfrac{-3+4i}{2+i}$

\[\Rightarrow \dfrac{{{z}_{1}}{{z}_{2}}}{{{{\bar{z}}}_{1}}}=\dfrac{-3+4i}{2+i}\times \dfrac{2-i}{2-i}\]

\[\Rightarrow \dfrac{{{z}_{1}}{{z}_{2}}}{{{{\bar{z}}}_{1}}}=\dfrac{-6+3i+8i+4}{4+1}\]

\[\Rightarrow \dfrac{{{z}_{1}}{{z}_{2}}}{{{{\bar{z}}}_{1}}}=\dfrac{-2+11i}{5}\].

Thus, \[\operatorname{Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{{{{\bar{z}}}_{1}}} \right)=-\dfrac{2}{5}\].

11. Express in the form of $a+ib:\left( 3i-7 \right)+(7-4i)-(6+3i)+{{i}^{23}}$.

Ans:

Let us simplify the given expression –

$\left( 3i-7 \right)+\left( 7-4i \right)-\left( 6+3i \right)+{{i}^{23}}=3i-7+7-4i-6-3i+{{i}^{23}}$

\[\Rightarrow -4i-6+{{i}^{22+1}}\]

\[\Rightarrow -4i-6+{{({{i}^{2}})}^{11}}.i\]

\[\Rightarrow -4i-6+{{(-1)}^{11}}.i\]

\[\Rightarrow -4i-6-i\]

Therefore, $\left( 3i-7 \right)+\left( 7-4i \right)-\left( 6+3i \right)+{{i}^{23}}=-6-5i$.

12. Find the conjugate of $\sqrt{-3}+4{{i}^{2}}$.

Ans:

Let us simplify the given expression –

\[\sqrt{-3}+4{{i}^{2}}=\sqrt{3}i-4\]

Therefore, the conjugate will be \[\bar{z}=-\sqrt{3}i-4\].

13. Solve for $x$ and $y$, $3x+(2x-y)i=6-3i$.

Ans:

We will equate the real part of the right-hand side with the real part of the left-hand side. Similarly, we will equate their imaginary parts as well.

Therefore,

\[3x=6\] and

\[2x-y=-3\]

\[\Rightarrow x=2\] and

\[\Rightarrow 2(2)-y=-3\]

\[\Rightarrow 4-y=-3\]

\[\Rightarrow y=7\].

Hence, \[x=2\] and 

\[y=7\].

14. Find the value of $1+{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+{{i}^{8}}+......+{{i}^{20}}$.

Ans:

Let us simplify the given expression –

\[1+{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+{{i}^{8}}+......+{{i}^{20}}=1-1+{{({{i}^{2}})}^{2}}+{{({{i}^{2}})}^{3}}+{{({{i}^{2}})}^{4}}+......+{{({{i}^{2}})}^{10}}\]

\[= {{(-1)}^{2}}+{{(-1)}^{3}}+{{(-1)}^{4}}+......+{{(-1)}^{10}}\]

\[= 1-1+1-1+......+1\]

\[= 1\]

Therefore, \[1+{{i}^{2}}+{{i}^{4}}+{{i}^{6}}+{{i}^{8}}+......+{{i}^{20}}=1\].

15. Multiply $3-2i$ by its conjugate.

Ans:

Let there be a complex number \[z=3-2i\]

Hence, its conjugate will be \[\bar{z}=3+2i\].

Therefore, the product of the complex number with its conjugate will be –

\[z\bar{z}=(3-2i)(3+2i)\]

\[\Rightarrow 9+6i-6i-4{{i}^{2}}\]

\[\Rightarrow 9+4\]

Hence, \[(3-2i)(3+2i)=13\].

16. Find the multiplicative inverse of \[4-3i\].

Ans:

Let us assume a complex number \[z=4-3i\].

To find the multiplicative inverse, we need \[\bar{z}\] and \[\left| z \right|\].

Therefore,

\[\bar{z}=4+3i\].

\[\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\]

\[\Rightarrow \sqrt{{{4}^{2}}+{{(-3)}^{2}}}\]

\[\Rightarrow \sqrt{16+9}\]

\[\Rightarrow \left| z \right|=5\]

Hence, multiplicative inverse \[{{z}^{-1}}=\dfrac{{\bar{z}}}{{{\left| z \right|}^{2}}}\] will be –

\[{{z}^{-1}}=\dfrac{4+3i}{25}\].

\[\Rightarrow {{z}^{-1}}=\dfrac{4}{25}+i\dfrac{3}{25}\].

17.   Express in the form of\[a+ib:\frac{\left( 3+i\sqrt{5} \right)\left( 3-i\sqrt{5} \right)}{\left( \sqrt{3}+i\sqrt{2} \right)-\left( \sqrt{3}-i\sqrt{2} \right)}\]

Ans:

Let us simplify the given expression –

\[\dfrac{\left( 3+i\sqrt{5} \right)\left( 3-i\sqrt{5} \right)}{\left( \sqrt{3}+i\sqrt{2} \right)-\left( \sqrt{3}-i\sqrt{2} \right)}=\dfrac{3(3-i\sqrt{5})+i\sqrt{5}(3-i\sqrt{5})}{\sqrt{3}+i\sqrt{2}-\sqrt{3}+i\sqrt{2}}\]

\[= \dfrac{9+5}{i2\sqrt{2}}\]

\[= \dfrac{14}{i2\sqrt{2}}\]

\[= \dfrac{7}{i\sqrt{2}}\times \dfrac{i\sqrt{2}}{i\sqrt{2}}\]

\[= -i\dfrac{7\sqrt{2}}{2}\]

Hence, \[\dfrac{\left( 3+i\sqrt{5} \right)\left( 3-i\sqrt{5} \right)}{\left( \sqrt{3}+i\sqrt{2} \right)-\left( \sqrt{3}-i\sqrt{2} \right)}=-i\dfrac{7\sqrt{2}}{2}\].

18. Evaluate the expression ${{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}$.

Ans:

Let us simplify the given expression –

\[{{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}={{i}^{n}}+{{i}^{n}}.i+{{i}^{n}}.{{i}^{2}}+{{i}^{n}}.{{i}^{2}}.i\]

\[= {{i}^{n}}+{{i}^{n}}.i+(-1){{i}^{n}}+(-1){{i}^{n}}.i\]

\[= {{i}^{n}}(1+i-1-i)\]

\[= 0\]

Therefore, \[{{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}=0\]

19. If $1,w,{{w}^{2}}$ are three cube root of unity, show that $(1-w+{{w}^{2}})(1+w-{{w}^{2}})=4$.

Ans:

Let us simplify the given expression –

\[(1-w+{{w}^{2}})(1+w-{{w}^{2}})=(1+w-{{w}^{2}})-w(1+w-{{w}^{2}})+{{w}^{2}}(1+w-{{w}^{2}})\]

\[= (1+w-{{w}^{2}})-w-{{w}^{2}}+{{w}^{3}}+{{w}^{2}}+{{w}^{3}}-{{w}^{4}})\]

\[= 1-{{w}^{2}}+{{w}^{3}}+{{w}^{3}}-{{w}^{4}}\]

Since, \[1,w,{{w}^{2}}\] are three cube roots of unity implies \[{{w}^{3}}=1\].

Therefore,

\[1-{{w}^{2}}+1+1-{{w}^{3}}.w\]

\[\Rightarrow 3-{{w}^{2}}-w\]

Now, we know that $w+{{w}^{2}}=-1$.

Hence, \[3-({{w}^{2}}+w)\]

\[\Rightarrow 3+1=4\]

Hence, proved.

20. Find the sum and product of the complex numbers $-\sqrt{3}+\sqrt{-2}$ and $2\sqrt{3}-i$.

Ans:

Given we have two complex numbers as ${{z}_{1}}=-\sqrt{3}+\sqrt{-2}$ and

${{z}_{2}}=2\sqrt{3}-i$.

Sum will be –

${{z}_{1}}+{{z}_{2}}=-\sqrt{3}+\sqrt{-2}+2\sqrt{3}-i$

$= -\sqrt{3}+2\sqrt{3}+\sqrt{2}i-i$

$= \sqrt{3}+i(\sqrt{2}-1)$

Product will be –

${{z}_{1}}{{z}_{2}}=(-\sqrt{3}+\sqrt{-2})(2\sqrt{3}-i)$

$= -2{{(\sqrt{3})}^{2}}+\sqrt{3}i+2\sqrt{6}i-\sqrt{2}{{i}^{2}}$

$= (\sqrt{2}-6)+\sqrt{3}i(1+2\sqrt{2})$.

21. Write the real and imaginary part $1-2{{i}^{2}}$.

Ans:

Let us simplify the given expression –

$1-2{{i}^{2}}=1-2(-1)$

$= 1+2$

$= 3$

Hence, the real and imaginary parts of $1-2{{i}^{2}}$ are $3$ and $0$.

22. If two complex numbers \[{{z}_{1}},{{z}_{2}}\] are such that \[\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|\], is it then necessary that \[{{z}_{1}}={{z}_{2}}\].

Ans:

Let us consider ${{z}_{1}}=x+iy$ and

${{z}_{2}}=y+ix$.

Hence, $\left| {{z}_{1}} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and

$\left| {{z}_{2}} \right|=\sqrt{{{y}^{2}}+{{x}^{2}}}$.

Therefore, it is not necessary that if two complex numbers ${{z}_{1}},{{z}_{2}}$ are such that $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|$ then ${{z}_{1}}={{z}_{2}}$.

23. Find the conjugate and modulus of \[\overline{9-i}+\overline{6+{{i}^{3}}}-\overline{9+{{i}^{2}}}\].

Ans:

Let us simplify the given expression –

$\overline{9-i}+\overline{6+{{i}^{3}}}-\overline{9+{{i}^{2}}}=9+i+\overline{6+{{i}^{2}}.i}-\overline{9-1}$

$= 9+i+\overline{6-i}-\overline{8}$

$= 9+i+6+i-8$

$= 7+2i$

Hence, the conjugate will be $\bar{z}=7-2i$.

Modulus will be $\left| z \right|=\sqrt{49+4}$

$\left| z \right|=\sqrt{53}$.

24. Find the number of non-zero integral solutions of the equation \[{{\left| 1-i \right|}^{x}}={{2}^{x}}\].

Ans:

Let us simplify the given expression –

${{\left| 1-i \right|}^{x}}={{2}^{x}}$

$\Rightarrow {{\left( \sqrt{{{1}^{2}}+{{(-1)}^{2}}} \right)}^{x}}={{2}^{x}}$

\[\Rightarrow {{\left( \sqrt{2} \right)}^{x}}={{2}^{x}}\]

\[\Rightarrow {{2}^{\dfrac{x}{2}}}={{2}^{x}}\]

Hence, \[\dfrac{x}{2}=x\]

\[\Rightarrow 2x-x=0\]

\[\Rightarrow x=0\].

Therefore, the integral solution of the equation ${{\left| 1-i \right|}^{x}}={{2}^{x}}$ is 

\[x=0\]. Hence, the number of non-zero integral solutions will be zero.

25. If \[(a+ib)(c+id)(e+if)(g+ih)=A+iB\], then show that: \[({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})({{e}^{2}}+{{f}^{2}})({{g}^{2}}+{{h}^{2}})={{A}^{2}}+{{B}^{2}}\].

Ans:

We know that \[\left| {{z}_{1}}{{z}_{2}} \right|=\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|\].

Therefore,

$\Rightarrow \left| (a+ib)(c+id)(e+if)(g+ih) \right|=\left| A+iB \right|$

$\Rightarrow \left| (a+ib) \right|\left| (c+id) \right|\left| (e+if) \right|\left| (g+ih) \right|=\left| A+iB \right|$

\[\Rightarrow \sqrt{{{a}^{2}}+{{b}^{2}}}\times \sqrt{{{c}^{2}}+{{d}^{2}}}\times \sqrt{{{e}^{2}}+{{f}^{2}}}\times \sqrt{{{g}^{2}}+{{h}^{2}}}=\sqrt{{{A}^{2}}+{{B}^{2}}}\]

\[\Rightarrow \sqrt{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)\left( {{e}^{2}}+{{f}^{2}} \right)\left( {{g}^{2}}+{{h}^{2}} \right)}=\sqrt{{{A}^{2}}+{{B}^{2}}}\]

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

Hence, proved.

Long Answer Questions: (4 Marks)

1. If \[x+iy=\dfrac{a+ib}{a-ib}\], Prove that \[{{x}^{2}}+{{y}^{2}}=1\].

Ans:

We have \[x+iy=\dfrac{a+ib}{a-ib}\].

Let us rationalise the denominator.

Hence, \[\dfrac{a+ib}{a-ib}=\dfrac{a+ib}{a-ib}\times \dfrac{a+ib}{a+ib}\]

\[\Rightarrow \dfrac{a+ib}{a-ib}=\dfrac{{{\left( a+ib \right)}^{2}}}{{{a}^{2}}-{{i}^{2}}{{b}^{2}}}\]

\[\Rightarrow \dfrac{a+ib}{a-ib}=\dfrac{{{a}^{2}}-{{b}^{2}}+2abi}{{{a}^{2}}+{{b}^{2}}}\]

\[\Rightarrow x+iy=\dfrac{{{a}^{2}}-{{b}^{2}}+2abi}{{{a}^{2}}+{{b}^{2}}}\]

Now, we will equate both real and imaginary parts from both sides.

Therefore,

\[x=\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] and

\[y=\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}}\]

Hence, we have –

\[{{x}^{2}}+{{y}^{2}}=\dfrac{{{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}}{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}}+\dfrac{4{{a}^{2}}{{b}^{2}}}{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{{{a}^{4}}+{{b}^{4}}-2{{a}^{2}}{{b}^{2}}+4{{a}^{2}}{{b}^{2}}}{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{{{a}^{4}}+{{b}^{4}}+2{{a}^{2}}{{b}^{2}}}{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}}{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}=1\]

Hence, proved.

2. Find real \[\theta \] such that \[\dfrac{3+2i\sin \theta }{1-2i\sin \theta }\] is purely real.

Ans:

Let us rationalise the denominator.

Hence, 

\[\dfrac{3+2i\sin \theta }{1-2i\sin \theta }=\dfrac{3+2i\sin \theta }{1-2i\sin \theta }\times \dfrac{1+2i\sin \theta }{1+2i\sin \theta }\]

\[\Rightarrow \dfrac{3+2i\sin \theta }{1-2i\sin \theta }=\dfrac{3+6i\sin \theta +2i\sin \theta +4{{i}^{2}}{{\sin }^{2}}\theta }{1-4{{i}^{2}}{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{3+2i\sin \theta }{1-2i\sin \theta }=\dfrac{3+8i\sin \theta -4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{3+2i\sin \theta }{1-2i\sin \theta }=\dfrac{3-4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+i\dfrac{8\sin \theta }{1+4{{\sin }^{2}}\theta }\]

For the above expression to be purely real the imaginary part should be equal to zero.

Therefore,

\[i\dfrac{8\sin \theta }{1+4{{\sin }^{2}}\theta }=0\]

\[\Rightarrow 8\sin \theta =0\]

\[\Rightarrow \theta ={{\sin }^{-1}}(\sin n\pi )\]

\[\Rightarrow \theta =n\pi \].

3. Find the modulus of \[\dfrac{(1+i)(2+i)}{3+i}\].

Ans:

Let us rationalise the denominator.

Hence, 

\[\dfrac{(1+i)(2+i)}{3+i}=\dfrac{2+i+2i-1}{3+i}\times \dfrac{3-i}{3-i}\]

\[\Rightarrow \dfrac{(1+i)(2+i)}{3+i}=\dfrac{(1+3i)(3-i)}{9+1}\]

\[\Rightarrow \dfrac{(1+i)(2+i)}{3+i}=\dfrac{3-i+9i+3}{10}\]

\[\Rightarrow \dfrac{(1+i)(2+i)}{3+i}=\dfrac{6+8i}{10}\]

\[\Rightarrow \dfrac{(1+i)(2+i)}{3+i}=\dfrac{3+4i}{5}\]

Therefore, the modulus will be –

\[\Rightarrow \left| z \right|=\sqrt{{{\left( \dfrac{3}{5} \right)}^{2}}+{{\left( \dfrac{4}{5} \right)}^{2}}}\]

\[\Rightarrow \left| z \right|=\sqrt{\dfrac{9}{25}+\dfrac{16}{25}}\]

\[\Rightarrow \left| z \right|=1\].

4. If \[\left| a+ib \right|=1\] then Show that \[\dfrac{1+b+ia}{1+b-ia}=b+ia\].

Ans:

Given we have \[\left| a+ib \right|=1\].

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=1\].

Now, let us simplify the left-hand side –

\[\dfrac{1+b+ia}{1+b-ia}=\dfrac{(1+b)+ia}{(1+b)-ia}\times \dfrac{(1+b)+ia}{(1+b)+ia}\]

\[\Rightarrow \dfrac{1+b+ia}{1+b-ia}=\dfrac{{{\left[ (1+b)+ia \right]}^{2}}}{{{(1+b)}^{2}}+{{a}^{2}}}\]

\[\Rightarrow \dfrac{1+b+ia}{1+b-ia}=\dfrac{{{(1+b)}^{2}}+2ia(1+b)-{{a}^{2}}}{{{(1+b)}^{2}}+{{a}^{2}}}\]

\[\Rightarrow \dfrac{1+b+ia}{1+b-ia}=\dfrac{1+2b+{{b}^{2}}+2ia+2iab-{{a}^{2}}}{1+2b+{{b}^{2}}+{{a}^{2}}}\]

Therefore,

\[\dfrac{1+b+ia}{1+b-ia}=\dfrac{{{a}^{2}}+{{b}^{2}}+2b+{{b}^{2}}+2ia+2iab-{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}+2b+{{b}^{2}}+{{a}^{2}}}\]

\[\Rightarrow \dfrac{1+b+ia}{1+b-ia}=\dfrac{2{{b}^{2}}+2b+2ia+2iab}{2{{a}^{2}}+2{{b}^{2}}+2b}\]

\[\Rightarrow \dfrac{1+b+ia}{1+b-ia}=\dfrac{{{b}^{2}}+b+ia+iab}{{{a}^{2}}+{{b}^{2}}+b}\]

\[\Rightarrow \dfrac{1+b+ia}{1+b-ia}=\dfrac{b(b+1)+ia(1+b)}{1+b}\]

\[\Rightarrow \dfrac{1+b+ia}{1+b-ia}=b+ia\].

Hence, proved.

5.   If $x-iy=\sqrt{\dfrac{a-ib}{c-id}}$ prove that 

${{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}$.

Ans:

We have \[x-iy=\sqrt{\dfrac{a-ib}{c-id}}\].

Let us rationalise the denominator.

Hence, \[\sqrt{\dfrac{a-ib}{c-id}}=\sqrt{\dfrac{a-ib}{c-id}\times \dfrac{c+id}{c+id}}\]

\[\Rightarrow \sqrt{\dfrac{a-ib}{c-id}}=\sqrt{\dfrac{ac+iad-ibc+bd}{{{c}^{2}}+{{d}^{2}}}}\]

\[\Rightarrow \sqrt{\dfrac{a-ib}{c-id}}=\sqrt{\dfrac{ac+bd+i(ad-bc)}{{{c}^{2}}+{{d}^{2}}}}\]

Therefore,

\[x-iy=\sqrt{\dfrac{ac+bd+i(ad-bc)}{{{c}^{2}}+{{d}^{2}}}}\]

\[\Rightarrow {{\left( x-iy \right)}^{2}}=\dfrac{ac+bd+i(ad-bc)}{{{c}^{2}}+{{d}^{2}}}\]

\[\Rightarrow {{x}^{2}}-{{y}^{2}}-2ixy=\dfrac{ac+bd+i(ad-bc)}{{{c}^{2}}+{{d}^{2}}}\]

By comparing both the sides, we get –

\[{{x}^{2}}-{{y}^{2}}=\dfrac{ac+bd}{{{c}^{2}}+{{d}^{2}}}\] and

\[-2xy=\dfrac{(ad-bc)}{{{c}^{2}}+{{d}^{2}}}\].

Now, we know that \[{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{\left( {{x}^{2}}-{{y}^{2}} \right)}^{2}}+4{{x}^{2}}{{y}^{2}}\].

\[\Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{\left( \dfrac{ac+bd}{{{c}^{2}}+{{d}^{2}}} \right)}^{2}}+{{\left( \dfrac{ad-bc}{{{c}^{2}}+{{d}^{2}}} \right)}^{2}}\]

\[\Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=\dfrac{{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}+2abcd-2abcd}{{{\left( {{c}^{2}}+{{d}^{2}} \right)}^{2}}}\]

\[\Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=\dfrac{{{a}^{2}}({{c}^{2}}+{{d}^{2}})+{{b}^{2}}({{c}^{2}}+{{d}^{2}})}{{{\left( {{c}^{2}}+{{d}^{2}} \right)}^{2}}}\]

\[\Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=\dfrac{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{c}^{2}}+{{d}^{2}} \right)}{{{\left( {{c}^{2}}+{{d}^{2}} \right)}^{2}}}\]

\[\Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{c}^{2}}+{{d}^{2}}}\].

Hence, proved.

6.   If \[a+ib=\dfrac{c+i}{c-i}\], where \[a,b,c\] are real. Prove that \[{{a}^{2}}+{{b}^{2}}=1\] and \[\dfrac{b}{a}=\dfrac{2c}{{{c}^{2}}-1}\].

Ans:

Given we have \[a+ib=\dfrac{c+i}{c-i}\].

Let us simplify the right-hand side –

Hence,

\[\dfrac{c+i}{c-i}=\dfrac{c+i}{c-i}\times \dfrac{c+i}{c+i}\]

\[\Rightarrow \dfrac{c+i}{c-i}=\dfrac{{{\left( c+i \right)}^{2}}}{{{c}^{2}}+1}\]

\[\Rightarrow \dfrac{c+i}{c-i}=\dfrac{{{c}^{2}}+2ic-1}{{{c}^{2}}+1}\]

\[\Rightarrow a+ib=\dfrac{{{c}^{2}}-1}{{{c}^{2}}+1}+i\dfrac{2c}{{{c}^{2}}+1}\]

By comparing real and imaginary parts we get –

\[a=\dfrac{{{c}^{2}}-1}{{{c}^{2}}+1}\] and

\[b=\dfrac{2c}{{{c}^{2}}+1}\]

Therefore,

\[{{a}^{2}}+{{b}^{2}}={{\left( \dfrac{{{c}^{2}}-1}{{{c}^{2}}+1} \right)}^{2}}+{{\left( \dfrac{2c}{{{c}^{2}}+1} \right)}^{2}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=\dfrac{{{\left( {{c}^{2}}-1 \right)}^{2}}}{{{\left( {{c}^{2}}+1 \right)}^{2}}}+\dfrac{4{{c}^{2}}}{{{\left( {{c}^{2}}+1 \right)}^{2}}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=\dfrac{{{\left( {{c}^{2}}+1 \right)}^{2}}}{{{\left( {{c}^{2}}+1 \right)}^{2}}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=1\].

Hence, proved.

Similarly, we have –

\[\dfrac{b}{a}=\dfrac{\dfrac{2c}{{{c}^{2}}+1}}{\dfrac{{{c}^{2}}-1}{{{c}^{2}}+1}}\]

\[\Rightarrow \dfrac{b}{a}=\dfrac{2c}{{{c}^{2}}-1}\]

Hence, proved.

7.   If \[{{z}_{1}}=2-i,\] 

\[{{z}_{2}}=1+i,\] find \[\left| \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+i} \right|\].

Ans:

Given we have ${{z}_{1}}=2-i$ and

${{z}_{2}}=1+i$.

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

$\Rightarrow {{z}_{1}}+{{z}_{2}}+1=4$

Also, ${{z}_{1}}-{{z}_{2}}+i=2-i-(1+i)+i$

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

$\Rightarrow {{z}_{1}}-{{z}_{2}}+i=1-i$.

Hence, the equation \[\left| \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+i} \right|\] becomes –

\[\left| \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+i} \right|=\left| \dfrac{4}{1-i} \right|\]

\[\Rightarrow \left| \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+i} \right|=\dfrac{\left| 4 \right|}{\left| 1-i \right|}\]

\[\Rightarrow \left| \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+i} \right|=\dfrac{4}{\sqrt{2}}\]

\[\Rightarrow \left| \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+i} \right|=\dfrac{4}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}\]

\[\Rightarrow \left| \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+i} \right|=\dfrac{4\sqrt{2}}{2}\]

Hence, the value of \[\left| \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+i} \right|=2\sqrt{2}\].

8.   If \[{{(p+iq)}^{2}}=x+iy\]. Prove that \[{{({{p}^{2}}+{{q}^{2}})}^{2}}={{x}^{2}}+{{y}^{2}}\].

Ans:

Let us simplify the left-hand side –

\[{{(p+iq)}^{2}}={{p}^{2}}-{{q}^{2}}+2pqi\]

\[\Rightarrow {{p}^{2}}-{{q}^{2}}+2pqi=x+iy\]

Therefore, after comparing real and imaginary parts of both sides, we get –

\[{{p}^{2}}-{{q}^{2}}=x\] and

\[2pq=y\].

Hence,

\[{{\left( {{p}^{2}}-{{q}^{2}} \right)}^{2}}+4{{p}^{2}}{{q}^{2}}={{x}^{2}}+{{y}^{2}}\]

\[\Rightarrow {{p}^{4}}+{{q}^{4}}-2{{p}^{2}}{{q}^{2}}+4{{p}^{2}}{{q}^{2}}={{x}^{2}}+{{y}^{2}}\]

\[\Rightarrow {{p}^{4}}+{{q}^{4}}+2{{p}^{2}}{{q}^{2}}={{x}^{2}}+{{y}^{2}}\]

\[{{\left( {{p}^{2}}+{{q}^{2}} \right)}^{2}}={{x}^{2}}+{{y}^{2}}\].

Hence, proved.

9.   If \[a+ib=\dfrac{{{(x+i)}^{2}}}{2{{x}^{2}}+1}\]. Prove that \[{{a}^{2}}+{{b}^{2}}=\dfrac{{{({{x}^{2}}+1)}^{2}}}{{{(2{{x}^{2}}+1)}^{2}}}\].

Ans:

Let us simplify the right-hand side of the given expression –

\[\dfrac{{{(x+i)}^{2}}}{2{{x}^{2}}+1}=\dfrac{{{x}^{2}}-1+2xi}{2{{x}^{2}}+1}\]

\[\Rightarrow a+ib=\dfrac{{{x}^{2}}-1+2xi}{2{{x}^{2}}+1}\]

Comparing real and imaginary parts of both sides, we get –

\[a=\dfrac{{{x}^{2}}-1}{2{{x}^{2}}+1}\] and

\[b=\dfrac{2x}{2{{x}^{2}}+1}\].

Hence,

\[{{a}^{2}}+{{b}^{2}}={{\left( \dfrac{{{x}^{2}}-1}{2{{x}^{2}}+1} \right)}^{2}}+{{\left( \dfrac{2x}{2{{x}^{2}}+1} \right)}^{2}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=\dfrac{{{\left( {{x}^{2}}-1 \right)}^{2}}}{{{\left( 2{{x}^{2}}+1 \right)}^{2}}}+\dfrac{4{{x}^{2}}}{{{\left( 2{{x}^{2}}+1 \right)}^{2}}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=\dfrac{{{x}^{4}}+1+2{{x}^{2}}}{{{\left( 2{{x}^{2}}+1 \right)}^{2}}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=\dfrac{{{({{x}^{2}}+1)}^{2}}}{{{\left( 2{{x}^{2}}+1 \right)}^{2}}}\]

Hence, proved.

10. If \[{{(x+iy)}^{3}}=u+iv\], then show that: 

\[\dfrac{u}{x}+\dfrac{v}{y}=4({{x}^{2}}-{{y}^{2}})\].

Ans:

We will simplify the given equation by using the identity \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\].

Therefore,

\[{{\left( x+iy \right)}^{3}}={{x}^{3}}+{{i}^{3}}{{y}^{3}}+3xyi\left( x+iy \right)\]

\[\Rightarrow {{x}^{3}}-i{{y}^{3}}+3{{x}^{2}}yi-3x{{y}^{2}}\]

\[\Rightarrow x\left( {{x}^{2}}-3{{y}^{2}} \right)+iy\left( 3{{x}^{2}}-{{y}^{2}} \right)\]

Hence, \[x\left( {{x}^{2}}-3{{y}^{2}} \right)+iy\left( 3{{x}^{2}}-{{y}^{2}} \right)=u+iv\].

Comparing both real and imaginary parts, we get –

\[u=x\left( {{x}^{2}}-3{{y}^{2}} \right)\] and

\[v=y\left( 3{{x}^{2}}-{{y}^{2}} \right)\].

Therefore, $\dfrac{u}{x}+\dfrac{v}{y}=\left( {{x}^{2}}-3{{y}^{2}} \right)+\left( 3{{x}^{2}}-{{y}^{2}} \right)$

$\Rightarrow \dfrac{u}{x}+\dfrac{v}{y}=4{{x}^{2}}-4{{y}^{2}}$

$\Rightarrow \dfrac{u}{x}+\dfrac{v}{y}=4\left( {{x}^{2}}-{{y}^{2}} \right)$.

Hence, proved.

11. Solve \[\sqrt{3}{{x}^{2}}-\sqrt{2}x+3\sqrt{3}=0\].

Ans:

As, the given equation is a quadratic polynomial. Therefore, we can find the roots of the equation by determining the discriminant.

Hence, we have $a=\sqrt{3}$,

$b=-\sqrt{2}$, and

$c=3\sqrt{3}$.

Therefore, roots will be –

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$\Rightarrow x=\dfrac{\sqrt{2}\pm \sqrt{2-36}}{2\sqrt{3}}$

$\Rightarrow x=\dfrac{\sqrt{2}\pm \sqrt{34}\times \sqrt{-1}}{2\sqrt{3}}$

$\Rightarrow x=\dfrac{\sqrt{2}\pm \sqrt{34}i}{2\sqrt{3}}$.

12. Find the modulus of \[{{i}^{25}}+{{(1+3i)}^{3}}\].

Ans:

Let us simplify the given expression –

${{i}^{25}}+{{(1+3i)}^{3}}={{({{i}^{2}})}^{12}}.i+1+27{{i}^{3}}+9i-27$

$\Rightarrow i+1-27i+9i-27$

$\Rightarrow z=-17i-26$.

Therefore, the modulus will be –

$\left| z \right|=\sqrt{{{\left( 17 \right)}^{2}}+{{\left( 26 \right)}^{2}}}$

$\left| z \right|=\sqrt{965}$

13. If \[a+ib=\dfrac{{{(x+i)}^{2}}}{2x-i}\]. Prove that \[{{a}^{2}}+{{b}^{2}}=\dfrac{{{({{x}^{2}}+1)}^{2}}}{4{{x}^{2}}+1}\].

Ans:

Let us simplify the right-hand side of the given expression –

\[\dfrac{{{(x+i)}^{2}}}{2x-i}=\dfrac{{{x}^{2}}-1+2xi}{2x-i}\]

\[\Rightarrow \dfrac{{{(x+i)}^{2}}}{2x-i}=\dfrac{{{x}^{2}}-1+2xi}{2x-i}\times \dfrac{2x+i}{2x+i}\]

\[\Rightarrow \dfrac{{{(x+i)}^{2}}}{2x-i}=\dfrac{2{{x}^{3}}+i{{x}^{2}}-2x-i+4{{x}^{2}}i-2x}{4{{x}^{2}}+1}\]

\[\Rightarrow \dfrac{{{(x+i)}^{2}}}{2x-i}=\dfrac{2{{x}^{3}}-4x+i(5{{x}^{2}}-1)}{4{{x}^{2}}+1}\]

\[\Rightarrow a+ib=\dfrac{2{{x}^{3}}-4x+i(5{{x}^{2}}-1)}{4{{x}^{2}}+1}\]

Comparing real and imaginary parts of both sides, we get –

\[a=\dfrac{2{{x}^{3}}-4x}{4{{x}^{2}}+1}\] and

\[b=\dfrac{5{{x}^{2}}-1}{4{{x}^{2}}+1}\].

Hence,

\[{{a}^{2}}+{{b}^{2}}={{\left( \dfrac{2{{x}^{3}}-4x}{4{{x}^{2}}+1} \right)}^{2}}+{{\left( \dfrac{5{{x}^{2}}-1}{4{{x}^{2}}+1} \right)}^{2}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=\dfrac{{{\left( 2{{x}^{3}}-4x \right)}^{2}}+{{\left( 5{{x}^{2}}-1 \right)}^{2}}}{{{\left( 4{{x}^{2}}+1 \right)}^{2}}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=\dfrac{4{{x}^{6}}+16{{x}^{2}}-16{{x}^{4}}+25{{x}^{4}}+1-10{{x}^{2}}}{{{\left( 4{{x}^{2}}+1 \right)}^{2}}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=\dfrac{4{{x}^{6}}+6{{x}^{2}}+9{{x}^{4}}+1}{{{\left( 4{{x}^{2}}+1 \right)}^{2}}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=\dfrac{\left( 4{{x}^{2}}+1 \right)\left( {{x}^{4}}+2{{x}^{2}}+1 \right)}{{{\left( 4{{x}^{2}}+1 \right)}^{2}}}\]

\[\Rightarrow {{a}^{2}}+{{b}^{2}}=\dfrac{{{\left( {{x}^{2}}+1 \right)}^{2}}}{4{{x}^{2}}+1}\]

Hence, proved.

14. Evaluate the given expression: ${{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}$.

Ans:

Let us simplify the given expression –

${{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}={{\left[ {{\left( {{i}^{2}} \right)}^{9}}+{{\left( \dfrac{1}{i} \right)}^{24+1}} \right]}^{3}}$

\[\Rightarrow {{\left[ {{\left( -1 \right)}^{9}}+\dfrac{1}{{{i}^{24}}.i} \right]}^{3}}\]

\[\Rightarrow {{\left[ -1+\dfrac{1}{{{\left( {{i}^{2}} \right)}^{12}}.i} \right]}^{3}}\]

\[\Rightarrow {{\left[ -1+\dfrac{1}{i} \right]}^{3}}\]

\[\Rightarrow -{{\left[ 1-\dfrac{1}{i} \right]}^{3}}\]

Now, we will further solve by using the identity \[{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)\].

\[\Rightarrow -{{\left[ 1-\dfrac{1}{i} \right]}^{3}}=-\left[ {{1}^{3}}-{{\left( \dfrac{1}{i} \right)}^{3}}-3\times \dfrac{1}{i}\left( 1-\dfrac{1}{i} \right) \right]\]

\[\Rightarrow -\left[ 1-\dfrac{1}{{{i}^{2}}.i}-\dfrac{3}{i}+\dfrac{3}{{{i}^{2}}} \right]\]

\[\Rightarrow -\left[ 1+\dfrac{1}{i}-\dfrac{3}{i}-3 \right]\]

\[\Rightarrow -\left[ -2-\dfrac{2}{i}\times \dfrac{i}{i} \right]\]

\[\Rightarrow -\left[ -2+2i \right]\]

Hence, \[{{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}=2-2i\].

15. Find the modulus and argument of \[\dfrac{1+i}{1-i}\].

Ans:

Let us simplify the given expression –

\[\dfrac{1+i}{1-i}=\dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}\]

\[\Rightarrow \dfrac{1+i}{1-i}=\dfrac{1+2i-1}{1+1}\]

\[\Rightarrow \dfrac{1+i}{1-i}=\dfrac{2i}{2}\]

\[\Rightarrow z=i\]

Hence, the modulus will be –

\[\left| z \right|=\sqrt{{{0}^{2}}+{{1}^{2}}}\]

\[\Rightarrow \left| z \right|=1\].

Argument will be –

$\theta ={{\tan }^{-1}}\dfrac{y}{x}$

$\Rightarrow \theta ={{\tan }^{-1}}\dfrac{1}{0}$

$\Rightarrow \theta ={{\tan }^{-1}}(\tan \dfrac{\pi }{2})$

$\Rightarrow \theta =\dfrac{\pi }{2}$.

Hence, the modulus is \[1\] and argument is $\dfrac{\pi }{2}$.

16. For what real value of \[x\] and \[y\] are numbers equal \[(1+i){{y}^{2}}+(6+i)\] and \[(2+i)x\].

Ans:

Let us equate both the numbers as –

\[(1+i){{y}^{2}}+(6+i)=(2+i)x\]

\[\Rightarrow ({{y}^{2}}+6)+i({{y}^{2}}+1)=2x+ix\]

\[\Rightarrow {{y}^{2}}+6=2x\] and

\[{{y}^{2}}+1=x\]

After solving both the equations obtained, we get –

\[x=5\] and

\[y=\pm 2\].

17. If \[x+iy=\sqrt{\dfrac{1+i}{1-i}}\], prove that \[{{x}^{2}}+{{y}^{2}}=1\].

Ans:

Let us simplify the right-hand side of the given expression –

\[\sqrt{\dfrac{1+i}{1-i}}=\sqrt{\dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}}\]

\[\Rightarrow \sqrt{\dfrac{1+i}{1-i}}=\sqrt{\dfrac{{{\left( 1+i \right)}^{2}}}{1+1}}\]

\[\Rightarrow \sqrt{\dfrac{1+i}{1-i}}=\dfrac{1+i}{\sqrt{2}}\]

\[\Rightarrow x+iy=\dfrac{1+i}{\sqrt{2}}\]

Comparing real and imaginary parts of both sides, we get –

\[x=\dfrac{1}{\sqrt{2}}\] and

\[y=\dfrac{1}{\sqrt{2}}\].

Hence,

\[{{x}^{2}}+{{y}^{2}}={{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}=\dfrac{1}{2}+\dfrac{1}{2}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}=1\]

Hence, proved.

18. Convert in the polar form $\dfrac{1+7i}{{{(2-i)}^{2}}}$.

Ans:

Let us simplify the given expression –

$\dfrac{1+7i}{{{(2-i)}^{2}}}=\dfrac{1+7i}{4-1-4i}$

$\Rightarrow \dfrac{1+7i}{3-4i}\times \dfrac{3+4i}{3+4i}$

$\Rightarrow \dfrac{3+4i+21i-28}{9+16}$

$\Rightarrow \dfrac{25i-25}{25}$

$\Rightarrow z=-1+i$.

Now, let $r\cos \theta =-1$ and 

$r\sin \theta =1$.

Now, let us add the squared value of both the equations.

Therefore,

${{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta =2$

$\Rightarrow {{r}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=2$

$\Rightarrow r=\sqrt{2}$.

Hence, we have –

$\sqrt{2}\cos \theta =-1$ and 

$\sqrt{2}\sin \theta =1$.

$\Rightarrow \cos \theta =-\dfrac{1}{\sqrt{2}}$ and

$\Rightarrow \sin \theta =\dfrac{1}{\sqrt{2}}$.

Now, since we have positive value for the sine function and negative for cosine function. Hence, the $\theta $ lies in the second quadrant.

$\Rightarrow \theta =\pi -\dfrac{\pi }{4}$

$\Rightarrow \theta =\dfrac{3\pi }{4}$

Hence, the polar form will be –

$-1+i=r(\cos \theta +i\sin \theta )$

$\Rightarrow -1+i=\sqrt{2}\left[ \cos \dfrac{3\pi }{4}+i\sin \dfrac{3\pi }{4} \right]$.

19. Find the real numbers \[x\] and \[y\] if \[(x-iy)(3+5i)\] is the conjugate of \[-6-24i\].

Ans:

We know that the conjugate of the given complex number $z=-6-24i$ will be $\bar{z}=-6+24i$.

Now, let us simplify the expression $(x-iy)(3+5i)$-

Hence, $(x-iy)(3+5i)=3x+5xi-3yi+5y$

$\Rightarrow (3x+5y)+i(5x-3y)$

Now, we will compare the values of the expression with the conjugate of the complex number.

Therefore,

$-6+24i=(3x+5y)+i(5x-3y)$

Thus, we have –

$-6=3x+5y$ and 

$24=5x-3y$.

Now, we will solve both equations for $x$ and $y$.

Hence, we have $51=-17y$

$\Rightarrow y=-3$

After substituting the value in first equation we get –

$\Rightarrow x=3$.

Therefore, the real numbers $x$ and $y$ are $3$ and $-3$.

20. If \[\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=1\], prove that \[\left| \dfrac{1}{{{z}_{1}}}+\dfrac{1}{{{z}_{2}}} \right|=\left| {{z}_{1}}+{{z}_{2}} \right|\].

Ans:

Given we have \[\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=1\]

\[\Rightarrow {{\left| {{z}_{1}} \right|}^{2}}={{\left| {{z}_{2}} \right|}^{2}}=1\]

\[\Rightarrow {{z}_{1}}\overline{{{z}_{1}}}=1\]

\[\Rightarrow \overline{{{z}_{1}}}=\dfrac{1}{{{z}_{1}}}\]

Similarly,

\[{{z}_{2}}\overline{{{z}_{2}}}=1\]

\[\Rightarrow \overline{{{z}_{2}}}=\dfrac{1}{{{z}_{2}}}\]

Therefore, we have –

\[\left| \dfrac{1}{{{z}_{1}}}+\dfrac{1}{{{z}_{2}}} \right|=\left| \overline{{{z}_{1}}}+\overline{{{z}_{2}}} \right|\]

\[\Rightarrow \left| \dfrac{1}{{{z}_{1}}}+\dfrac{1}{{{z}_{2}}} \right|=\left| \overline{{{z}_{1}}+{{z}_{2}}} \right|\]

\[\Rightarrow \left| \dfrac{1}{{{z}_{1}}}+\dfrac{1}{{{z}_{2}}} \right|=\left| {{z}_{1}}+{{z}_{2}} \right|\]

Hence, proved.

Long Answer Questions: (6 Marks)

1. If \[z=x+iy\] and \[w=\dfrac{1-iz}{z-i}\]. Show that \[\left| w \right|=1\Rightarrow z\] is purely real.

Ans:

Given we have \[z=x+iy\].

Hence, we have –

\[w=\dfrac{1-i(x+iy)}{x+iy-i}\]

\[\Rightarrow w=\dfrac{1+y-ix}{x+i(y-1)}\]

As, \[\left| w \right|=1\]

Hence,

\[\left| \dfrac{1+y-ix}{x+i(y-1)} \right|=1\]

\[\Rightarrow \dfrac{\left| 1+y-ix \right|}{\left| x+i(y-1) \right|}=1\]

\[\Rightarrow \dfrac{\sqrt{{{(1+y)}^{2}}+{{x}^{2}}}}{\sqrt{{{x}^{2}}+{{(y-1)}^{2}}}}=1\]

\[\Rightarrow {{(1+y)}^{2}}+{{x}^{2}}={{x}^{2}}+{{(y-1)}^{2}}\]

\[\Rightarrow 1+{{y}^{2}}+2y={{y}^{2}}+1-2y\]

\[\Rightarrow 4y=0\]

Hence, \[z=x+0i\] implying that \[z\] is purely real.

2. Convert into polar form \[\dfrac{-16}{1+i\sqrt{3}}\].

Ans:

Let us simplify the given expression –

$\dfrac{-16}{1+i\sqrt{3}}=\dfrac{-16}{1+i\sqrt{3}}\times \dfrac{1-i\sqrt{3}}{1-i\sqrt{3}}$

$\Rightarrow \dfrac{-16}{1+i\sqrt{3}}=\dfrac{-16(1-i\sqrt{3})}{1+3}$

$\Rightarrow \dfrac{-16}{1+i\sqrt{3}}=-4+i4\sqrt{3}$

Now, let $r\cos \theta =-4$ and 

$r\sin \theta =4\sqrt{3}$.

Now, let us add the squared value of both the equations.

Therefore,

${{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta =16+48$

$\Rightarrow {{r}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=64$

$\Rightarrow r=8$.

Hence, we have –

$8\cos \theta =-4$ and 

$8\sin \theta =4\sqrt{3}$.

$\Rightarrow \cos \theta =-\dfrac{1}{2}$ and

$\Rightarrow \sin \theta =\dfrac{\sqrt{3}}{2}$.

Now, since we have positive values for the sine function and negative for the cosine function. Hence, the $\theta $ lies in the second quadrant.

$\Rightarrow \theta =\pi -\dfrac{\pi }{3}$

$\Rightarrow \theta =\dfrac{2\pi }{3}$

Hence, the polar form will be –

$-4+i4\sqrt{3}=r(\cos \theta +i\sin \theta )$

$\Rightarrow -4+i4\sqrt{3}=8\left[ \cos \dfrac{2\pi }{3}+i\sin \dfrac{2\pi }{3} \right]$.

3. Find two numbers such that their sum is \[6\] and the product is \[14\].

Ans:

Let us consider $x$ and $y$ as the two numbers.

Hence, we have –

$x+y=6$ and

$xy=14$.

Therefore,

$x(6-x)=14$

$\Rightarrow 6x-{{x}^{2}}=14$

$\Rightarrow {{x}^{2}}-6x+14=0$

$\Rightarrow x=\dfrac{6\pm \sqrt{36-56}}{2}$

$\Rightarrow x=\dfrac{6\pm \sqrt{20}i}{2}$

$\Rightarrow x=3\pm \sqrt{5}i$

Hence, $y=6-(3\pm \sqrt{5}i)$

$\Rightarrow y=3\mp \sqrt{5}i$

4. Convert into polar form \[z=\dfrac{i-1}{\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3}}\].

Ans:

Let us simplify the given expression –

$\dfrac{i-1}{\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3}}=\dfrac{i-1}{\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}}$

$\Rightarrow \dfrac{i-1}{\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3}}=\dfrac{2(i-1)}{1+i\sqrt{3}}$

$\Rightarrow \dfrac{i-1}{\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3}}=\dfrac{2i-2}{1+i\sqrt{3}}\times \dfrac{1-i\sqrt{3}}{1-i\sqrt{3}}$

$\Rightarrow \dfrac{i-1}{\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3}}=\dfrac{2i+2\sqrt{3}-2+i2\sqrt{3}}{1+3}$

$\Rightarrow \dfrac{i-1}{\cos \dfrac{\pi }{3}+i\sin \dfrac{\pi }{3}}=\dfrac{\sqrt{3}-1+i(\sqrt{3}+1)}{2}$

Now, let \[r\cos \theta =\dfrac{\sqrt{3}-1}{2}\] and 

$r\sin \theta =\dfrac{\sqrt{3}+1}{2}$.

Now, let us add the squared value of both the equations.

Therefore,

${{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta =\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{4}+\dfrac{{{\left( \sqrt{3}+1 \right)}^{2}}}{4}$

$\Rightarrow {{r}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )=\dfrac{4-2\sqrt{3}+4+2\sqrt{3}}{4}$

$\Rightarrow r=2$.

Hence, we have –

$2\cos \theta =\dfrac{\sqrt{3}-1}{2}$ and 

$2\sin \theta =\dfrac{\sqrt{3}+1}{2}$.

$\Rightarrow \cos \theta =\dfrac{\sqrt{3}-1}{4}$ and

$\Rightarrow \sin \theta =\dfrac{\sqrt{3}+1}{4}$.

Now, since we have positive value for both the functions. Hence, the $\theta $ lies in the first quadrant.

$\Rightarrow \theta =\dfrac{\pi }{4}+\dfrac{\pi }{6}$

$\Rightarrow \theta =\dfrac{5\pi }{12}$

Hence, the polar form will be –

$\dfrac{\sqrt{3}-1+i(\sqrt{3}+1)}{2}=r(\cos \theta +i\sin \theta )$

$\Rightarrow \dfrac{\sqrt{3}-1+i(\sqrt{3}+1)}{2}=2\left[ \cos \dfrac{5\pi }{12}+i\sin \dfrac{5\pi }{12} \right]$.

5. If \[\alpha \] and \[\beta \] are different complex numbers with \[\left| \beta  \right|=1\], then find \[\left| \dfrac{\beta -\alpha }{1-\bar{\alpha }\beta } \right|\].

Ans:

Let us assume that $\alpha ={{a}_{1}}+i{{b}_{1}}$ and

$\beta ={{a}_{2}}+i{{b}_{2}}$.

Hence, $\bar{\alpha }={{a}_{1}}-i{{b}_{1}}$

Now, it is given that $\left| \beta  \right|=1$.

Also, $\left| \beta  \right|=\sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}$

$\Rightarrow \sqrt{{{a}_{2}}^{2}+{{b}_{2}}^{2}}=1$

\[\Rightarrow {{a}_{2}}^{2}+{{b}_{2}}^{2}=1\ \ ......\text{(1)}\]

Now, we need to find $\left| \dfrac{\beta -\alpha }{1-\bar{\alpha }\beta } \right|$.

Therefore,

$\left| \dfrac{\beta -\alpha }{1-\bar{\alpha }\beta } \right|=\left| \dfrac{{{a}_{2}}+i{{b}_{2}}-{{a}_{1}}-i{{b}_{1}}}{1-\left( {{a}_{1}}-i{{b}_{1}} \right)\left( {{a}_{2}}+i{{b}_{2}} \right)} \right|$

$\Rightarrow \left| \dfrac{{{a}_{2}}-{{a}_{1}}+i\left( {{b}_{2}}-{{b}_{1}} \right)}{1-\left( {{a}_{1}}{{a}_{2}}+i{{b}_{2}}{{a}_{1}}-i{{b}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)} \right|$

$\Rightarrow \left| \dfrac{{{a}_{2}}-{{a}_{1}}+i\left( {{b}_{2}}-{{b}_{1}} \right)}{1-{{a}_{1}}{{a}_{2}}-{{b}_{1}}{{b}_{2}}-i\left( {{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}} \right)} \right|$

Now, we know that \[\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\dfrac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}\].

Hence,

$\Rightarrow \dfrac{\left| {{a}_{2}}-{{a}_{1}}+i\left( {{b}_{2}}-{{b}_{1}} \right) \right|}{\left| 1-{{a}_{1}}{{a}_{2}}-{{b}_{1}}{{b}_{2}}-i\left( {{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}} \right) \right|}$

$\Rightarrow \dfrac{\sqrt{{{\left( {{a}_{2}}-{{a}_{1}} \right)}^{2}}+{{\left( {{b}_{2}}-{{b}_{1}} \right)}^{2}}}}{\sqrt{{{\left( 1-{{a}_{1}}{{a}_{2}}-{{b}_{1}}{{b}_{2}} \right)}^{2}}+{{\left( {{b}_{2}}{{a}_{1}}-{{b}_{1}}{{a}_{2}} \right)}^{2}}}}$

\[\Rightarrow \dfrac{\sqrt{{{a}_{2}}^{2}+{{a}_{1}}^{2}-2{{a}_{1}}{{a}_{2}}+{{b}_{2}}^{2}+{{b}_{1}}^{2}-2{{b}_{1}}{{b}_{2}}}}{\sqrt{1+{{\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)}^{2}}-2\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)+{{b}_{2}}^{2}{{a}_{1}}^{2}+{{b}_{1}}^{2}{{a}_{2}}^{2}-2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}}}\]

From equation \[\text{(1)}\], we have \[{{a}_{2}}^{2}+{{b}_{2}}^{2}=1\].

Therefore,

\[\Rightarrow \dfrac{\sqrt{1+{{a}_{1}}^{2}-2({{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}})+{{b}_{1}}^{2}}}{\sqrt{1+{{a}_{1}}^{2}{{a}_{2}}^{2}+{{b}_{1}}^{2}{{b}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}-2\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)+{{b}_{2}}^{2}{{a}_{1}}^{2}+{{b}_{1}}^{2}{{a}_{2}}^{2}-2{{a}_{1}}{{a}_{2}}{{b}_{1}}{{b}_{2}}}}\]

\[\Rightarrow \dfrac{\sqrt{1+{{a}_{1}}^{2}-2({{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}})+{{b}_{1}}^{2}}}{\sqrt{1+{{a}_{1}}^{2}({{a}_{2}}^{2}+{{b}_{2}}^{2})+{{b}_{1}}^{2}({{a}_{2}}^{2}+{{b}_{2}}^{2})-2\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)}}\]

\[\Rightarrow \dfrac{\sqrt{1+{{a}_{1}}^{2}+{{b}_{1}}^{2}-2({{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}})}}{\sqrt{1+{{a}_{1}}^{2}+{{b}_{1}}^{2}-2\left( {{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}} \right)}}\]

\[\Rightarrow 1\]

Hence, $\left| \dfrac{\beta -\alpha }{1-\bar{\alpha }\beta } \right|=1$.

CBSE Class 11 Maths Chapter-5 Important Questions - Free PDF Download

Class 11 Maths Chapter 5 Important Questions

Vedantu provides important questions for class 11 maths complex numbers. The topic of Chapter 5 of Mathematics is Complex Numbers. The chapter Complex numbers and Quadratic equations class 11 covers how to solve sums based on complex numbers.

Topics that are discussed below are:

• The real number and an imaginary number

• Complex number definition

• Integral powers of iota

• Purely real and purely imaginary complex numbers.

• Complex number equality

• Algebra in complex number – addition, subtraction, multiplication, and division

• Conjugate of a complex number.

• Modulus in a complex number

• Argand plane

• Polar form

Before knowing the concept of the complex number, one must know the meaning of the real number and imaginary number.

Important Definitions

Real Number: A number on the number line that is in the form of a positive, negative, rational, irrational, zero, fraction, integer, etc is called a real number. 

For example: 34, -3, \[\sqrt{7}\], 22/67, 0 

Imaginary Number: The numbers except the real numbers is called an imaginary number. It is the root of a negative number.

For example: \[\sqrt{-67}\], \[\sqrt{-23/92}\], etc.

Complex Number: A complex number is defined as the number that can be expressed in the form of a + ib.

Here, a and b are real numbers and i is iota which will be discussed.

The value of iota is R-1. 

Therefore, z (complex number) = a + ib where a is the real part, and ib is the imaginary part. 

a = Re (z)

b= Im (z)

We can differentiate between the real part and the imaginary part of a complex number with the help of a table given below.

The Integral Powers of Lota 

i =  \[\sqrt{-1}\]

i² = -1 

i³ = i².i = (-1).i = -i

i⁴ = i².i² = (-1) (-1) = 1

Now we can generalise it, therefore;

i\[^{4n+1}\] = i

i\[^{4n+2}\] = = -1

i\[^{4n+3}\] = -i

i\[^{4n+4}\] = 1

It can also be generalized in this form,

If n= even integer then, i\[^{n}\] = (-1)\[^{n/2}\]

And, if n= odd integer then, i\[^{n}\] = (-1)\[^{(n-1)/2}\]

What is a purely real complex number and a purely imaginary complex number?

If the imaginary part of a complex number is zero, then it is called a purely real complex number.

Im(z) = 0

If the real part of a complex number is zero, then it is called a purely imaginary complex number.

Re(z) = 0

Complex Number Equality

Two complex numbers are taken, z\[_{1}\] and z\[_{2}\]

z\[_{1}\] = a\[_{1}\] + ib\[_{1}\]

z\[_{2}\] = a\[_{2}\] + ib\[_{2}\]

If z\[_{1}\] = z\[_{2}\]

i.e. a\[_{1}\] + ib\[_{1}\] = a\[_{2}\] + ib\[_{2}\]

then, a\[_{1}\] = a\[_{2}\] and b\[_{1}\] = b\[_{2}\]

Therefore, Re(z\[_{1}\]) = Re(z\[_{2}\])

And Im(z\[_{1}\]) = Im(z\[_{2}\])

Algebra in Complex Number

a. Addition

Let, z\[_{1}\] = a\[_{1}\] + ib\[_{1}\]

and, z\[_{2}\] = a\[_{2}\] + ib\[_{2}\]

Adding both the complex numbers we get,

z\[_{1}\] + z\[_{2}\] = (a\[_{1}\] + ib\[_{1}\]) + (a\[_{2}\] + ib\[_{2}\])

=  (a\[_{1}\] + a\[_{2}\]) + i(b\[_{1}\] + b\[_{2}\])  

Properties:

1. Commutative: z\[_{1}\] + z\[_{2}\] = z\[_{2}\] + z\[_{1}\]  

2. Associative: (z\[_{1}\] + z\[_{2}\]) + z\[_{3}\] = z\[_{1}\] + (z\[_{2}\] + z\[_{3}\])

3. Additive identity: 0+z = z+0 = z

b. Subtraction

Let z\[_{1}\] = a\[_{1}\] + ib\[_{1}\] 

and, z\[_{2}\] = a\[_{2}\] + ib\[_{2}\]  

Subtracting the complex numbers we get:

z\[_{1}\] -  z\[_{2}\] = (a\[_{1}\] + ib\[_{1}\]) - (a\[_{2}\] + ib\[_{2}\])

= (a\[_{1}\] - a\[_{2}\]) + i(b\[_{1}\] - b\[_{2}\])

c. Multiplication 

Let z\[_{1}\] = a\[_{1}\] + ib\[_{1}\]

and, z\[_{2}\] = a\[_{2}\] + ib\[_{2}\] 

Multiplying both the complex numbers, we get

z\[_{1}\] z\[_{2}\]  = (a\[_{1}\] + ib\[_{1}\]) (a\[_{2}\] + ib\[_{2}\])

= (a\[_{1}\]a\[_{2}\] - b\[_{1}\]b\[_{2}\]) + i(a\[_{1}\]b\[_{2}\] + a\[_{2}\]b\[_{1}\])  

Properties:

1. Commutative: z\[_{1}\] z\[_{2}\] = z\[_{2}\] z\[_{1}\]

2. Associative: (z\[_{1}\]z\[_{2}\])z\[_{3}\] = z\[_{1}\](z\[_{2}\] z\[_{3}\])

3. Distributive law: z\[_{1}\](z\[_{2}\] + z\[_{3}\]) =  z\[_{1}\] z\[_{2}\] +  z\[_{1}\] z\[_{3}\]  

3. Multiplicative identity: z.1= 1.z

d. Division

Let z\[_{1}\] = a\[_{1}\] + ib\[_{1}\]

and, z\[_{2}\] = a\[_{2}\] + ib\[_{2}\] 

Dividing z\[_{1}\] by z\[_{2}\] , we get

\[\frac{z_{1}}{z_{2}}\] = \[\frac{(a_{1} + ib_{1})}{(a_{2} + ib_{2})}\]

=  \[\frac{(a_{1}a_{2} - b_{1}b_{2}) + i(a_{1}b_{2} + a_{2}b_{1})}{a_{1}^{2} + a_{2}^{2}}\],  z\[_{2}\] ≠ 0 

Solved Examples on Complex Numbers

1. Express (5 – 3i)³ in the form a + ib.

Solution:  We have, (5 – 3i)³ 

= 5³ – 3 × 5²× (3i) + 3 × 5 (3i)² – (3i)³ 

= 125 – 225i – 135 + 27i 

= – 10 – 198i

2. Simplify

a) 16i + 10i(3-i)

b) (7i)(5i)

c) 11i + 13i – 2i

Solution: 

a) 16i + 10i(3-i)

= 16i + 10i(3) + 10i (-i)

= 16i +30i – 10 i2

= 46 i – 10 (-1)

= 46i + 10

b) (7i)(5i) = 35  i2 = 35 (-1) = -35

c) 11i + 13i – 2i = 22i

Download important questions for class 11 maths chapter 5

Conjugate of a Complex Number

It is denoted by \[\bar{z}\]

\[\bar{z}\] = a - ib

Modulus of a Complex Number

If z= a+ib 

Then, ।z। = a\[^{2}\] + b\[^{2}\]

Argand Plane and Polar Plane

A plane just like XY plane where the complex number a+ib has the coordinates, a and b is called Argand plane. It is also known as the Gaussian plane. 

Argument of a complex number, z is shown by arg(z)= θ = tan-1(a/b)

arg(z) can also be written as amp(z).                                                   

z is 2nπ + θ is the general value of arg(z).

and the length of OP = \[\sqrt{a^{2}+b^{2}}\]

(Image will be Uploaded Soon)

The principal values of the argument lies in the interval (- π, π].

(i) Given x> 0 and y > 0 then, arg (z) = 0

(ii) Given x < 0 and y> 0 then, arg (z) = π -0

(iii) Given x < 0 and y < 0 then, arg (z) = – (π – θ)

(iv) Given x> 0 and y < 0 then, arg (z) = -θ

Polar Form

If z = a + ib is a complex number, then z in polar form can be written as, 

z = |z| (cos θ + i sin θ) where, θ = arg (z)

If the general value of the argument is 0, then the polar form of z is

z = |z|[cos (2nπ + θ) + i sin (2nπ + θ)], where n is an integer.

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Practice Questions of CBSE Class 11 Maths Chapter 5

Some Practice Questions of CBSE Class 11 Maths Chapter 5 are as follows.

1. Find out the solution of the quadratic equation 2x2 + x + 1 = 0.

Answer: (-1 ± √7i) / 4

2. If z1 and z2 are two complex numbers then prove that Re(z1z2) = Rez1 Rez2– Imz1Imz2

3. Evaluate i98.

Answer: -1

4. Calculate the the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)]

Answer: 3

5. Write the complex number (9 – i) – ( –4 + i3) in the form a + ib.

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

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Preparation for any exam is incomplete without practice. Students are required to practice questions in order to perform well in examinations. In Vedantu you will get sufficient material to practice with. 

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