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

## Chapter : 5

## What Are Complex Numbers And Quadratic Equations ?

**To download the complete Syllabus (PDF File), Please fill & submit the form below.**

## 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

**Addition of two complex numbers**

Let z_{1} = a + ib and z_{2} = c + id be any two complex numbers.

Then z_{1} + z_{2} = (a + ib) + (c + id) = (a + c) + i(b + d)

**Difference of two complex numbers**

Given, any two complex number z_{1} and z_{2} , then difference,

z_{1} – z_{2} = z_{1} + (– z_{2}) ⇒ (a + ib) – (c + id)

= (a + ib) + (– c – id)

= (a – c) + i(b – d)

**Multiplication of two complex numbers**

Let z_{1} = a + ib and z_{2} = c + id be two complex numbers.

Then product z_{1} z_{2} = (a + ib)(c + id) = ac – bd + i(ad + bc)

**Division or two complex number**

Let any two complex numbers z_{1} and z_{2} ,

where z_{2} ≠ 0, then quotient

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

**Power of i**

We know that

i^{3} = i^{2} × i = (– 1)i = – i

i^{4} = i^{2} × i^{2} = – 1 × – 1 = 1

i^{5} = (i^{2}) ^{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

i^{4k} = 1, i^{4k + 1} = i, i^{4k + 2} = – 1, i^{4k + 3} = i

**The square roots of negative real numbers**

Note that i^{2} = –1 and (– i)^{2} = i^{2} = –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.

## Quadratic Equations

The solutions of the quadratic equation ax^{2}+ bx + c = 0 where a, b, c ∈ R a ≠ 0,

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

A polynomial equation of degree n has n roots.