# Complex Numbers And Quadratic Equations Class 11 Notes Mathematics Chapter 5 - CBSE

## What Are Complex Numbers And Quadratic Equations ?

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## Complex Number

A number of the form (a + ib) where a and b are real numbers is defined to be a complex numbers.

$$\text{e.g. : 2+3i, -1+}\sqrt{3}\space i$$

## Algrebra Of Complex Numbers

Let z1 = a + ib and z2 = c + id be any two complex numbers.

Then z1 + z2 = (a + ib) + (c + id) = (a + c) + i(b + d)

Difference of two complex numbers

Given, any two complex number z1 and z2 , then difference,

z1 – z2 = z1 + (– z2) ⇒ (a + ib) – (c + id)

= (a + ib) + (– c – id)

= (a – c) + i(b – d)

Multiplication of two complex numbers

Let z1 = a + ib and z2 = c + id be two complex numbers.

Then product z1 z2 = (a + ib)(c + id) = ac – bd + i(ad + bc)

Division or two complex number

Let any two complex numbers z1 and z2 ,

where z2 ≠ 0, then quotient

$$\frac{z_{1}}{z_{2}}= \frac{z_{1}.1}{z_{2}}$$

Power of i

We know that

i3 = i2 × i = (– 1)i = – i

i4 = i2 × i2 = – 1 × – 1 = 1

i5 = (i2) 2 × i = (– 1)2 i = i, etc.

$$\text{Also\space}i^{\normalsize-1}=\frac{1}{i}× \frac{i}{i}=\frac{i}{i^{2}} = -i\\i^{\normalsize-2}=\frac{1}{i^{2}}=-1$$

In general, for any integer

i4k = 1, i4k + 1 = i, i4k + 2 = – 1, i4k + 3 = i

The square roots of negative real numbers

Note that i2 = –1 and (– i)2 = i2 = –1.

Therefore, the square roots of –1 are i, – i.

However, by the symbol $$\sqrt{\normalsize-1}$$ , we should mean i only.

Identities

$$\centerdot\space\text{(z}_1 + \text{z}_2)^2 =\\ z_1^2 + z_2^2 + 2z_1z_2\\\centerdot\space(Z_{1} - Z_{2})^{2} =\\ Z_{1}^{2} - 2Z_{1}Z_{2} + Z_{2}^{2}\\\centerdot\space (z_{1} + z_{2})^{3} =\\ z_{1}^{3}+ 3z_{1}^{2}z_{2}+3Z_{1}Z_{2}^{2}+z_{2}^{3}\\\centerdot\space(Z_{1}-Z_{2})^{3} =\\ z_{3}^{3}- 3z_{1}^{2}z_{2} + 3z_{1}z_{2}^{2}- z_{2}^{2}\\\centerdot\space z_{1}^{2} - z_{2}^{2}=\\\text{(z}_{1} + z_{2})(z_{1}-z_{2})$$

## The Modulus And Conjugate Of The Complex Numbers

Let z = a + ib be a complex number. Then the modulus of z, denoted by |z|. If z = a + ib then conjugate is represented by

$$\bar{z} =\text{a + ib}\\\therefore\space|z| =\sqrt{a^{2} + b^{2}}\\\text{and}\space\bar{z} = a - ib\\\centerdot|z_{1} z_{2}| =|Z_{1}||Z_{2}|\\\centerdot\space\bigg|\frac{z_{1}}{z_{2}}\bigg|=\bigg|\frac{z_{1}}{z_{2}}\bigg|\\\text{provided}\space |z_{2}|\neq 0\\\centerdot\space Z_{1}Z_{2} =\bar{z}_{1}\bar{z}_{2}\\\centerdot\space \overline{z_{1}\pm z_{2}} = \bar{z}_{1} + \bar{z}_{2}\\\centerdot\space \bigg(\frac{\bar{z}_{1}}{\bar{z}_{2}}\bigg)=\frac{\bar{Z}_{1}}{\bar{Z}_{2}}\\\text{provided |z}_2| ≠ 0$$

## Argand Plane And Polar Representation

The polar form of the complex number z = x + iy is r(cos θ + i sin θ)

$$\text{where r =}\sqrt{x^{2} + y^{2}}\\\text{(the modulus of z) and}\\\text{cos}\space\theta = \frac{x}{r},\space \text{sin}\space\theta=\frac{y}{r}\\\text{(}\theta\space \text{is known as the argument of z.)}$$

The value of θ, such that – π ≤ θ ≤π is the principal argument of z.

$$\text{b}^{2} - 4ac \lt 0\space\text{are given by}\\ x =\frac{-b\pm\sqrt{b^{2} - 4ac}}{2a}$$