Cbse Class 11 Maths Notes Chapter 4 Complex Numbers And Quadratic Equations

Section–A (1 Mark Questions)

1. If a+2i and 3+bi are conjugate of each other then find value of a and b.

Ans. Let $z=a+2i$

$\overline{z}=a-2i$

But we are given that $\overline{z}=3+bi$

$$\therefore a-2i=3+bi$$

Comparing the real and imaginary parts, we get $a=3$ and $b=-2$.

2. The conjugate of the complex number ${\displaystyle \frac{1-i}{1+i}}$ is ______.

Ans. $$\text{ 2. }\begin{array}{rl}& \frac{1-i}{1+i}=\frac{1-i}{1+i}\times \frac{1-i}{1-i}\\ & =\frac{1+i^2-2i}{1-i^2}\\ & =\frac{1-1-2i}{1+1}=-i\end{array}$$

Hence, conjugate of $\frac{1-i}{1+i}$ is $0+i$.

3. If z=2+3i , then value of $|z-1+4i|$ is______.

Ans. Given that: $z=2+3i$

Then, $|z-1+4i|=|2+3i-1+4i|$

$$\begin{array}{rl}& =|1+7i|=\sqrt{1^2+7^2}\\ & =\sqrt{50}=5\sqrt{2}\end{array}$$

4. The sum of the series $i+i^2+i^3+\cdots$ up to 1000 terms is equal to ……….

Ans. $i+i^2+i^3+\dots .$. upto 1000 terms

$$=i+i^2+i^3+\dots +i^{1000}=0$$

Hence, the value of the filler is 0 .

5. Find the amplitude of $\sqrt{3}+i$ .

Ans. Let $z=\sqrt{3}+i$

Here, $z$ lies in first quadrant.

Therefore 

$\tan \theta =\left|\frac{\mathrm{Im}(z)}{\mathrm{Re}(z)}\right|=\left|\frac{1}{\sqrt{3}}\right|=\frac{1}{\sqrt{3}}$

$$\therefore \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\text{ i.e., }\theta =\frac{\pi }{6}\text{. }$$

Section–B (2 Marks Questions)

6. Show that $(z+3)(\overrightarrow{z}+3)$ is equivalent to ${|z+3|}^2$ .

Ans. Let $z=x+iy$

$$\begin{array}{rl}& \mathrm{So}(z+3)(\overline{z}+3)=(x+iy+3)(x-iy+3)\\ & =[(x+3)+iy][(x+3)-iy]\\ & =(x+3{)}^2-i^2{y}^2\\ & =(x+3{)}^2+y^2\\ & =|x+3+iy{|}^2=|z+3{|}^2.\end{array}$$

7. If ${\left({\displaystyle \frac{1+i}{1-i}}\right)}^x=1$ , then show that $x=4n$ , where $nϵN$.

Ans. Given that: ${\left(\frac{1+i}{1-i}\right)}^x=1$

$$\begin{array}{rl}& \Rightarrow {\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)}^x=1\\ & \Rightarrow {\left(\frac{1+i^2+2i}{1-i^2}\right)}^x=1\\ & \Rightarrow {\left(\frac{1-1+2i}{1+1}\right)}^x=1\\ & \Rightarrow {\left(\frac{2i}{2}\right)}^x=1\\ & \Rightarrow (i{)}^x=(i{)}^{4n}\\ & \Rightarrow x=4n,n\in N\end{array}$$

8. Find the value of p such that the difference of the roots of the equation $x^2-px+8=0$ is 2.

Ans. Let $\alpha ,\beta$ be the roots of the equation

$$x^2-px+8=0$$

Therefore $\alpha +\beta =p$ and $\alpha \beta =8$

Now, $\alpha -\beta =\pm \sqrt{(\alpha +\beta {)}^2-4\alpha \beta }$

Therefore, $2=\pm \sqrt{p^2-32}$

$$\begin{array}{rl}& \Rightarrow p^2-32=4\\ & \Rightarrow p=\pm 6.\end{array}$$

9. If $x+iy={\displaystyle \frac{a+ib}{a-ib}}$, prove that $x^2+y^2=1$.

Ans. We have,

$$\begin{array}{rl}& x+iy=\frac{(a+ib)(a+ib)}{(a-ib)(a+ib)}\\ & =\frac{a^2-b^2+2abi}{a^2+b^2}=\frac{a^2-b^2}{a^2+b^2}+\frac{2ab}{a^2+b^2}i\end{array}$$

So that, $x-iy=\frac{a^2-b^2}{a^2+b^2}-\frac{2ab}{a^2+b^2}i$

Therefore,

$$\begin{array}{rl}& x^2+y^2=(x+iy)(x-iy)\\ & =\frac{{(a^2-b^2)}^2}{{(a^2+b^2)}^2}+\frac{4a^2{b}^2}{{(a^2+b^2)}^2}=\frac{{(a^2+b^2)}^2}{{(a^2+b^2)}^2}=1\end{array}$$

10. Express $\frac{5+\sqrt{2}i}{1-\sqrt{2}i}$  in the form of.

Ans. $$\text{ 10. }\begin{array}{rl}& \frac{5+\sqrt{2}i}{1-\sqrt{2}i}\\ & =\frac{5+\sqrt{2}i}{1-\sqrt{2}i}\times \frac{1+\sqrt{2}i}{1+\sqrt{2}i}\\ & =\frac{5+5\sqrt{2}i+\sqrt{2}i-2}{1-(\sqrt{2}i{)}^2}=\frac{3+6\sqrt{2}i}{1+2}\\ & =\frac{3(1+2\sqrt{2}i)}{3}=1+2\sqrt{2}i.\end{array}$$

11. What is the conjugate of ${\displaystyle \frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}}$

Ans. Let, $z=\frac{\sqrt{5+12i}+\sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}}$

12. Express $z=\frac{1}{(2+i{)}^2}$ in standard form.

Ans. $$\begin{array}{rl}& \text{ 12. }\frac{1}{(2+i{)}^2}=\frac{1}{4+i^2+4i}\\ & =\frac{1}{4-1+4i}=\frac{1}{3+4i}\\ & =\frac{1}{3+4i}\times \frac{3-4i}{3-4i}\\ & =\frac{3-4i}{9-16i^2}=\frac{3-4i}{25}\end{array}$$

13. If $\frac{(1+i{)}^2}{2-i}=x+iy$, find x+y.

Ans. Given: $\frac{(1+i{)}^2}{2-i}=x+iy$

Upon expansion we get,

$$\begin{array}{rl}& \frac{1^2+i^2+2(1)(i)}{2-i}=x+iy\\ & \frac{1+(-1)+2i}{2-i}=x+iy\\ & \frac{2i}{2-i}=x+iy\end{array}$$

Now, let us multiply and divide by $(2+i)$, we get

$$\begin{array}{rl}& \frac{2i}{2-i}\times \frac{2+i}{2+i}=x+iy\\ & \frac{4i+2i^2}{2^2-i^2}=x+iy\\ & \frac{2(-1)+4i}{4-(-1)}=x+iy\\ & \frac{-2+4i}{5}=x+iy\end{array}$$

Let us equate real and imaginary parts on both sides we get,

$x=-\frac{2}{5}$ and $y=\frac{4}{5}$

So, $x+y=-\frac{2}{5}+\frac{4}{5}=\frac{2}{5}$.

PDF Summary - Class 11 Maths Complex Numbers and Quadratic Equations Notes (Chapter 4)

1. Definition

When a given number is in the form of $a+ib$, where $a,b\in R$ and $i=\sqrt{-1}$ it is called a complex number and such number is denoted by ‘$z$’.

$z=a+ib$

Where,

$a$= real part of complex number and,

$b$= imaginary part of complex number.

1.1 Conjugate of a Complex Number

Consider a complex number $z=a+ib$,

Then its conjugate is written as '$\overline{z}$'.

Whose value is defined as $\overline{z}=a-ib$.

2. Algebra of Complex Numbers

Let $z_1=a+ib$ and $z_2=c+id$ be two complex numbers where $\text{a},\text{b},\text{c},\text{d}\in \text{R}$ and $\text{i}=\sqrt{-1}$.

1. Addition :

$z_1+z_2=(a+bi)+(c+di)$

$=(a+c)+(b+d)i$

2. Subtraction :

$z_1-z_2=(a+bi)-(c+di)$

$=(a-c)+(b-d)$

3. Multiplication :

$z_1\cdot z_2=(a+bi)(c+di)$

$=a(c+di)+bi(c+di)$

$=ac+adi+bci+bd{i}^2$

$=ac-bd+(ad+bc)i$    $(\because i^2=-1)$

Note…

1. $\begin{array}{r}a+ib=c+id\\ \iff a=c\mathrm{\&}b=d\end{array}$

2. $i^{4k+r}=\{\begin{array}{cl}1;& r=0\\ i;& r=1\\ -1;& r=2\\ -i;& r=3\end{array}$

3. $\sqrt{b}\sqrt{a}=\sqrt{ba}$  is only possible if atleast one of either a or b is non-negative.

3. Argand Plane

Any complex number $z=a+ib$ can be represented by a unique point $P(a,b)$ in the argand plane.

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$P(a,b)$ represents the complex number $z=a+ib$.

3.1 Modulus and Argument of Complex Number

Consider a complex number $z=a+ib$.

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(i) Distance of $z$ from the origin is referred to as the modulus of complex number $z$.

It is represented by $\text{r}=|\text{z}|=\sqrt{{\text{a}}^2+{\text{b}}^2}$

(ii) Here, $\theta$ i.e.The angle made by ray OP with positive direction of the real axis is called the argument of $\mathbf{z}$.

Note.

${\text{z}}_1>{\text{z}}_2$ or ${\text{z}}_1<{\text{z}}_2$ has no meaning but $\left|{\text{z}}_1\right|>\left|{\text{z}}_2\right|$ or $\left|{\text{z}}_1\right|<\left|{\text{z}}_2\right|$ holds meaning.

3.2 Principal Argument

The argument ${}^{\prime }{\theta }^{\prime }$ of complex numbers $z=a+ib$ is called the principal argument of $z$if $-\pi <\theta ⩽\pi$.

Consider$\tan \alpha =\left|{\displaystyle \frac{b}{a}}\right|$, and $\theta$ be the $\arg \left(z\right)$.

i. 

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ii.

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iii.

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iv.

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In (iii) and (iv) the principal argument is given by $-\pi +\alpha$ and $-\alpha$respectively.

4. Polar Form

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$\begin{array}{l}a=r\cos \theta \mathrm{\&}b=r\sin \theta \\ \text{ where }r=|z|\text{ and }\theta =\arg (z)\\ \therefore z=a+ib\\ =r(\cos \theta +\mathrm{isin}\theta )\end{array}$

Note…

A complex number $z$ can also be represented as $z=r{e}^{i\theta }$, it is known as Euler’s form.

Where,

$\mathrm{r}=|\mathrm{Z}|\mathrm{\&}\theta =\arg (\mathrm{Z})$

5. Some Important Properties

1. $\overline{(\overline{z})}=z$

2. $z+\overline{z}=2\mathrm{Re}(z)$

3. $z-\overline{z}=2i\mathrm{Im}(z)$

4. $\overline{{\text{z}}_1+{\text{z}}_2}={\bar{\text{z}}}_1+{\bar{\text{z}}}_2$

5. $\overline{{\text{z}}_1{\text{z}}_2}={\bar{\text{z}}}_1.{\bar{\text{z}}}_2$

6. $|\text{z}|=0\Rightarrow \text{z}=0$

7. $z.\overline{z}=|z{|}^2$

8. $\left|z_1{z}_2\right|=\left|z_1\right|\left|z_2\right|;\left|{\displaystyle \frac{z_1}{z_2}}\right|={\displaystyle \frac{\left|z_1\right|}{\left|z_2\right|}}$

9. $|\bar{\text{z}}|=|\text{z}|=|-\text{z}|$

10. ${\left|z_1\pm z_2\right|}^2={\left|z_1\right|}^2+{\left|z_2\right|}^2\pm 2\mathrm{Re}\left(z_1{\overline{z}}_2\right)$

11. $\left|z_1+z_2\right|⩽\left|z_1\right|+\left|z_2\right|$         (Triangle Inequality)

12. $\left|{\text{z}}_1-{\text{z}}_2\right|⩾\left|\left|{\text{z}}_1\right|-\left|{\text{z}}_2\right|\right|$

13. ${\left|\text{a}{\text{z}}_1-\text{b}{\text{z}}_2\right|}^2+{\left|\text{b}{\text{z}}_1+\text{a}{\text{z}}_2\right|}^2=\left({\text{a}}^2+{\text{b}}^2\right)\left({\left|{\text{z}}_1\right|}^2+{\left|{\text{z}}_2\right|}^2\right)$

14. $\mathrm{amp}\left({\text{z}}_1\cdot {\text{z}}_2\right)=\mathrm{amp}{\text{z}}_1+\mathrm{amp}{\text{z}}_2+2\text{k}\pi ;\text{k}\in \text{I}$

15. $\mathrm{amp}\left({\displaystyle \frac{{\text{y}}_0}{{\text{y}}_1}}\right)=\mathrm{amp}{\text{z}}_1-\mathrm{amp}{\text{z}}_2+2\text{k}\pi ;\text{k}\in \text{I}$

16. $\mathrm{amp}\left(z^n\right)=n\mathrm{amp}(z)+2k\pi ;k\in I$

6. De-Moivre's Theorem

Statement: $\cos n\theta +i\sin n\theta$ is the value or one of the values of $(\cos \theta +i\sin \theta {)}^n$ according as if 'n' is integer or a rational number. The theorem is very useful in determining the roots of any complex quantity.

7. Cube Root of Unity

Roots of the equation ${\text{x}}^3=1$ are called cube roots of unity.$

Roots of the equation $x^3=1$ are called cube roots of unity.

$\begin{array}{l}{x}^3-1=0\\ (x-1)(x^2+x+1)=0\\ x=1\text{ or }{x}^2+x+1=0\end{array}$

i.e $x=\underset{\mathrm{w}}{\underbrace{{\displaystyle \frac{-1+\sqrt{3}\mathrm{i}}{2}}}}$ or $\mathrm{x}=\underset{w^2}{\underbrace{{\displaystyle \frac{-1-\sqrt{3}\mathrm{i}}{2}}}}$

(i) The cube roots of unity are $1,{\displaystyle \frac{-1+i\sqrt{3}}{2}},{\displaystyle \frac{-1-i\sqrt{3}}{2}}$.

(ii) ${\omega }^3=1$

(iii) If $\text{w}$ is one of the imaginary cube roots of unity then $1+\omega +{\omega }^2=0$.

(iv) In general $1+{\omega }^{\text{r}}+{\omega }^{2\text{r}}=0$; where $\text{r}\in \text{I}$ but is not the multiple of 3 .

(v) In polar form the cube roots of unity are:

$\cos 0+i\sin 0;\cos {\displaystyle \frac{2\pi }{3}}+i\sin {\displaystyle \frac{2\pi }{3}},\cos {\displaystyle \frac{4\pi }{3}}+i\sin {\displaystyle \frac{4\pi }{3}}$

(vi) The three cube roots of unity when plotted on the argand plane constitute the vertices of an equilateral triangle.

(vii) The following factorisation should be remembered:

$\begin{array}{l}{a}^3-b^3=(a-b)(a-\omega b)(a-{\omega }^{2}b)\\ x^2+x+1=(x-\omega )(x-{\omega }^2)\\ a^3+b^3=(a+b)(a+\omega b)(a+{\omega }^{2}b)\\ a^3+b^3+c^3-3abc=(a+b+c)(a+\omega b+{\omega }^{2}c)(a+{\omega }^{2}b+\omega c)\end{array}$

8. ‘n’ nth Roots of Unity

Solution of equation $x^n=1$ is given by,

$\begin{array}{ll}\text{x}=\cos {\displaystyle \frac{2\text{k}\pi }{\text{n}}}+\text{i}\sin {\displaystyle \frac{2\text{k}\pi }{\text{n}}}& ;\text{k}=0,1,2,\dots ,\text{n}-1\\ ={\text{e}}^{\text{i}\left({\displaystyle \frac{2\text{k}\pi }{\text{n}}}\right)}& ;\text{k}=0,1,\dots .,\text{n}-1\end{array}$

Note.

1. We may take any n consecutive integral values of $\text{k}$ to get ' $\text{n}$ ' ${\text{n}}^{\text{th }}$ roots of unity.

2. Sum of ' ${}^{\prime }{n}^{\prime }{n}^{\text{th }}$ roots of unity is zero, $n\in N$

3. The points represented by '$\text{n}$',  nth roots of unity are located at the vertices of a regular polygon of $n$ sides inscribed in a unit circle, centered at the origin and one vertex being one positive real axis.

Properties:

If $1,{\alpha }_1,{\alpha }_2,{\alpha }_3\dots .{\alpha }_{n-1}$ are the $\text{n},{\text{n}}^{\text{th }}$ root of unity then:

(i) They are in G.P. with common ratio ${\text{e}}^{\text{i}(2\pi /\text{n})}$

(ii) $1^p+{\alpha }_0^0+{\alpha }_1^{\circ }+\dots .+{\alpha }_{m-0}^{\circ }=[\begin{array}{l}0,\text{ if }p\ne k^n\\ n,\text{ if }p=kn\end{array}$ where $k\in Z$

(iii) $\left(1-{\alpha }_1\right)\left(1-{\alpha }_2\right)\dots \dots \left(1-{\alpha }_{n-1}\right)=\text{n}$

(iv) $\left(1+{\alpha }_1\right)\left(1+{\alpha }_2\right)\dots \dots \left(1+{\alpha }_{n-1}\right)=[\begin{array}{l}0,\text{ if n is even }\\ 1,\text{ if n is odd }\end{array}$

(v) $1.{\alpha }_1\cdot {\alpha }_2\cdot {\alpha }_3\dots \dots \dots {\alpha }_{n-1}=[\begin{array}{c}-1,\text{ if }n\text{ is even }\\ 1,\text{ if }n\text{ is odd }\end{array}$

Note...

(i) $\cos \theta +\cos 2\theta +\cos 3\theta +\dots .+\cos n\theta ={\displaystyle \frac{\sin (n\theta /2)}{\sin (\theta /2)}}\cos \left({\displaystyle \frac{n+1}{2}}\right)\theta$

(ii) $\sin \theta +\sin 2\theta +\sin 3\theta +\dots .+\sin n\theta ={\displaystyle \frac{\sin (n\theta /2)}{\sin (\theta /2)}}\sin \left({\displaystyle \frac{n+1}{2}}\right)\theta$.

9. Square Root of Complex Number

Let $\text{x}+\text{iy}=\sqrt{\text{a}+\text{ib}}$, Squaring both sides, we get

$(x+iy{)}^2=a+ib$

i.e., $x^2-y^2=a,2xy=b$

Solving these equations, we get square roots of $z$.

10. LOCI in Complex Plane

(i) $\left|z-z_0\right|=$ a represents the circumference of a circle, centred at ${\text{z}}_{\text{o}}$, radius a.

(ii) $\left|\text{z}-{\text{z}}_{\text{0}}\right|<$ a represents the interior of the circle.

(iii) $\left|z-z_0\right|>$ a represents the exterior of this circle.

(iv) $\left|z-z_1\right|=\left|z-z_2\right|$ represents $\mathrm{\perp }$ bisector of segment with endpoints $z_1\text{ and }{z}_2$.

(v) $\left|{\displaystyle \frac{-z_1}{-z_2}}\right|=k$ represents: $\left\{\begin{array}{l}\text{ circle, k}\ne 1\\ \mathrm{\perp }\text{ bisector, k}=1\end{array}\right\}$

(vi) $\arg (\text{z})=\theta$ is a ray starting from the origin (excluded) inclined at an $\mathrm{\angle }\theta$ with a real axis.

(vii) Circle described on line segment joining $z_1\text{ and }{z}_2$ as diameter is:

$\left(z-z_1\right)\left(\overline{z}-{\overline{z}}_2\right)+\left(z-z_2\right)\left(\overline{z}-{\overline{z}}_1\right)=0$

(viii) If ${\mathrm{z}}_1,{\mathrm{z}}_2,{\mathrm{z}}_3$ are the vertices of an equilateral triangle where ${\mathrm{z}}_0$ is its circumcentre then

(a) ${\displaystyle \frac{1}{{\mathrm{z}}_2-{\mathrm{z}}_3}}+{\displaystyle \frac{1}{{\mathrm{z}}_3-{\mathrm{z}}_1}}+{\displaystyle \frac{1}{{\mathrm{z}}_1-{\mathrm{z}}_2}}=0$

(b) $z_0^1+z_1^1+z_2^1-z_1{z}_2-z_2{z}_3-z_3{z}_1=0$

(c) $z_0^1+z_1^1+z_2^1=3z_{/}^1$

(ix) If $\mathrm{A},\mathrm{B},\mathrm{C}\mathrm{\&}\mathrm{D}$ are four points representing the complex numbers ${\mathrm{z}}_1,{\mathrm{z}}_2,{\mathrm{z}}_3\mathrm{\&}{\mathrm{z}}_4$ then

$\mathrm{AB}||\mathrm{CD}$ if ${\displaystyle \frac{{\mathrm{Z}}_4-{\mathrm{Z}}_3}{{\mathrm{z}}_2-{\mathrm{z}}_1}}$ is purely real ;

$\mathrm{AB}\perp \mathrm{CD}$ if ${\displaystyle \frac{{\mathrm{z}}_4-{\mathrm{z}}_3}{{\mathrm{z}}_2-{\mathrm{z}}_1}}$ is purely imaginary $]$

11. Vectorial Representation of a Complex

Every complex number can be considered as if it is the position vector of that point. If the point P represents the complex number z then,

$\overrightarrow{OP}=z\text{ and }\left|\overrightarrow{OP}\right|=\left|z\right|.$

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Note.

(i) If $\overrightarrow{\text{OP}}=\text{z}=\text{r}{\text{e}}^{\text{i}\theta }$ then $\overrightarrow{\text{OQ}}={\text{z}}_1=\text{r}{\text{e}}^{\text{i}(\theta +\varphi )}=\text{z}\text{.}{\text{e}}^{\text{i}\varphi }$.

If $\overrightarrow{\text{OP}}$ and $\overrightarrow{\text{OQ}}$ are of unequal magnitude, then

$\widehat{\text{OQ}}=\widehat{\text{OP}}{\text{e}}^{\text{i}\varphi }$

(ii) If $z_1,z_2\text{ and }{\text{z}}_3$ are three vertices of a triangle ABC described in the counterclockwise sense, then

${\displaystyle \frac{z_3-z}{z_2-z}}={\displaystyle \frac{AC}{AB}}(\cos \alpha +i\sin \alpha )={\displaystyle \frac{AC}{AB}}\cdot e^{i\alpha }={\displaystyle \frac{\left|z_3-z_1\right|}{\left|z_2-z_1\right|}}\cdot e^{i\alpha }$

12. Some Important Results

(i) If ${\text{z}}_1$ and ${\text{z}}_2$ are two complex numbers, then the distance between $z_1$ and $z_2$ is $\left|z_2-z_1\right|$.

(ii) Segment Joining points $\text{A}\left({\text{z}}_1\right)$ and $\text{B}\left({\text{z}}_2\right)$ is divided by point $\text{P}(\text{z})$ in the ratio ${\text{m}}_1:{\text{m}}_2$ then $\text{z}={\displaystyle \frac{{\text{m}}_1{\text{z}}_2+{\text{m}}_2\text{z}}{{\text{m}}_1+{\text{m}}_2}},{\text{m}}_1$ and ${\text{m}}_2$ are real.

(iii) The equation of the line joining $z_1$ and $z_2$ is given by

$\left|\begin{array}{cc}z& \overline{z}\\ z& \overline{z}\\ z_2& {\overline{z}}_2\end{array}\right|=0\text{ (non parametric form) }$

Or

${\displaystyle \frac{z-z}{\overline{z}-\overline{z}}}={\displaystyle \frac{z-z_2}{\overline{z}-{\overline{z}}_2}}$

(iv) $\overline{a}z+a\overline{z}+b=0$ represents a general form of line.

(v) The general eqn. of circle is:

$z\overline{z}+a\overline{z}+\overline{a}z+b=0$     (where $\text{b}$ is real no.).

Centre : $(-a)$ and radius,

$\sqrt{|a{|}^2-b}=\sqrt{a\overline{a}-b}$.

(vi) Circle described on line segment joining $z_1$ and $z_2$  as diameter is:

$\left(z-z_1\right)\left(\overline{z}-{\overline{z}}_2\right)+\left(z-z_2\right)\left(\overline{z}-{\overline{z}}_1\right)=0$

(vii) Four pts. $z_1,z_2,z_3,z_4$ in anticlockwise order will be concyclic, if and only if

$\theta =\arg \cdot \left({\displaystyle \frac{z_2-z_4}{z_1-z_4}}\right)=\arg \left({\displaystyle \frac{z_2-z_3}{z_1-z_3}}\right)$

$\Rightarrow \arg \left({\displaystyle \frac{z_2-z_4}{z_1-z_4}}\right)-\arg \cdot \left({\displaystyle \frac{z_2-z_3}{z_1-z_3}}\right)=2n\pi ;(n\in I)$

$\Rightarrow \arg \left[\left({\displaystyle \frac{z_2-z_4}{z_1-z_4}}\right)\left({\displaystyle \frac{z_1-z_3}{z_2-z_3}}\right)\right]=2n\pi$

$\Rightarrow \left({\displaystyle \frac{z_2-z_4}{z_1-z_4}}\right)\times \left({\displaystyle \frac{z_1-z_3}{z_2-z_3}}\right)$ is real and positive.

(viii) If $z_1,z_2,z_3$ are the vertices of an equilateral triangle where $z_0$ is its circumcentre then

(a) ${\displaystyle \frac{1}{z_2-z_3}}+{\displaystyle \frac{1}{z_3-z_1}}+{\displaystyle \frac{1}{z_1-z_2}}=0$

(b) $z_0^1+z_1^1+z_2^1-z_1{z}_2-z_2{z}_3-z_3{z}_1=0$

(c) ${\text{z}}_0^1+{\text{z}}_1^1+{\text{z}}_2^1=3{\text{z}}_{}^1$

(ix) If $A,B,C\text{ and }D$ are four points representing the complex numbers $z_1,z_2,z_3\text{ and }{z}_4$ then

$\text{AB}||\text{CD}$ if ${\displaystyle \frac{z_4-z_3}{z_2-z_1}}$ is purely real;

$AB\mathrm{\perp }CD\text{ if }{\displaystyle \frac{z_4-z_3}{z_2-z_1}}$ is purely imaginary.

(x) Two points $P\left(z_1\right)$ and $Q\left(z_2\right)$ lie on the same side or opposite side of the line $\overline{a}z+a\overline{z}+b$ accordingly as $\overline{a}{z}_1+a{\overline{z}}_1+b\text{ and }\overline{a}{z}_2+a{\overline{z}}_2+b$ have same sign or opposite sign.

Important Identities

(i) ${\text{x}}^2+\text{x}+1=(\text{x}-\omega )\left(\text{x}-{\omega }^2\right)$

(ii) ${\text{x}}^2-\text{x}+1=(\text{x}+\omega )\left(\text{x}+{\omega }^2\right)$

(iii) $x^2+xy+y^2=(x-y\omega )\left(x-y{\omega }^2\right)$

(iv) ${\text{x}}^2-\text{xy}+{\text{y}}^2=(\text{x}+\omega \text{y})\left(\text{x}+\text{y}{\omega }^2\right)$

(v) $x^2+y^2=(x+iy)(x-iy)$

(vi) $x^3+y^3=(x+y)(x+y\omega )\left(x+y{\omega }^2\right)$

(vii) ${\text{x}}^3-{\text{y}}^3=(\text{x}-\text{y})(\text{x}-\text{y}\omega )\left(\text{x}-\text{y}{\omega }^2\right)$

(viii) $x^2+y^2+z^2-xy-yz-zx=\left(x+y\omega +z{\omega }^2\right)\left(x+y{\omega }^2+z\omega \right)$

or $\left(\text{x}\omega +\text{y}{\omega }^2+\text{z}\right)\left(\text{x}{\omega }^2+\text{y}\omega +\text{z}\right)$

or $\left(\text{x}\omega +\text{y}+\text{z}{\omega }^2\right)\left(\text{x}{\omega }^2+\text{y}+\text{z}\omega \right)$

(ix) $x^3+y^3+z^3-3xyz=(x+y+z)\left(x+\omega y+{\omega }^{2}z\right)$$\left(\text{x}+{\omega }^2\text{y}+\omega \text{z}\right)$

1. Quadratic Expression

The standard form of a quadratic expression in $\text{x}$ is, $f(\text{x})=a{x}^2+bx+c$, where $a,b,c\in R\text{ and }a\ne 0$. General form of a quadratic equation in $\text{x}$ is, $a{x}^2+bx+c=0$, where $a,b,c\in R\text{ and }a\ne 0$.

2. Roots of Quadratic Equation

(a) The solution of the quadratic equation,

$a{x}^2+bx+c=0$ is given by $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

The expression $\text{D}={\text{b}}^2-4\text{ac}$ is called the discriminant of the quadratic equation.

(b) If $\alpha \text{ and }\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$, then

(i) $\alpha +\beta ={\displaystyle \frac{-\text{b}}{a}}$

(ii) $\alpha \beta ={\displaystyle \frac{\text{c}}{a}}$

(iii) $|\alpha -\beta |={\displaystyle \frac{\sqrt{\text{D}}}{|\text{a}|}}$

(c) A quadratic equation whose roots are $\alpha \text{ and }\beta$ is $(x-\alpha )(x-\beta )=0$ i.e.,

${\text{x}}^2-(\alpha +\beta )\text{x}+\alpha \beta =0$ i.e.,

${\text{x}}^2-($ sum of roots $)\text{x}+$ product of roots $=0$

Note.

$y=(a{x}^2+bx+c)\equiv a(x-\alpha )(x-\beta )$

$=a{(x+{\displaystyle \frac{b}{2a}})}^2-{\displaystyle \frac{D}{4a}}$

3. Nature of Roots

(a) Consider the quadratic equation $a{x}^2+bx+c=0$ where a, $a,b,c\in R\text{ and }a\ne 0$ then;

(i) $\text{D}>0\iff$ roots are real and distinct (unequal).

(ii) $\text{D}=0\iff$ roots are real and coincident (equal).

(iii) $\text{D}<0\iff$ roots are imaginary.

(iv) If $\text{p}+\text{iq}$ is one root of a quadratic equation, then the other must be the conjugate  $p-iq$ and vice versa. $(\text{p},\text{q}\in \text{R and i}=\sqrt{-1})$.

(b) Consider the quadratic equation $a{x}^2+bx+c=0$ where $a,b,c\in Q\text{ and }a\ne 0$ then;

(i) If $\text{D}>0$ and is a perfect square, then roots are rational and unequal.

(ii) If $\alpha =p+\sqrt{q}$ is one root in this case, (where $p$ is rational and $\sqrt{q}$ is a surd) then the other root must be the conjugate of it i.e., $\beta =p-\sqrt{q}$ and vice versa.

Note.

Remember that a quadratic equation cannot have three different roots and if it has, it becomes an identity.

4. Graph of Quadratic Equation

Consider the quadratic expression, $\text{y}=\text{a}{\text{x}}^2+\text{bx}+\text{c}$, $a\ne 0\text{ and }a,b,c\in R$ then;

(i) The graph between $\text{x},\text{y}$ is always a parabola. If $\text{a}>0$ then the shape of the parabola is concave upwards and if $\text{a}<0$ then the shape of the parabola is concave downwards.

(ii) $\text{y}>0\mathrm{\forall }\text{x}\in \text{R}$, only if $\text{a}>0\text{ and D}<0$