We know that the squares of real numbers is always non-negative, thus, there is no real number which can satisfy an equation of the type $x^{2}+k = 0$, where k > 0. in solving quadratic equations, we often get a square root of negative number as part of the values for roots.

The extraction of square root is not possible as no real numbers exists which when squared will yield a negative number.This shows the inadequacy of the real number system R. So, we need to widen the real number system so as to include square root of negative real number in such a manner that usual operations of addition, subtraction, multiplication and division and their related properties hold.

T
he new system is called the system of complex numbers and it include all the real numbers. In this section we will be learning more about complex numbers.

A number of the form $a+ib$, where $a$ and $b$ are real numbers and $i$ = $\sqrt{-1}$, is called a complex number. The complex number $a+ib$ is generally denoted by $z$.
Example : $2+3i$, $-3+5i$ are complex numbers.

The set of all complex numbers is denoted by $C$.

i.e., $C$ = ($a+ib: a, b \in R$ and $i= \sqrt{-1}$).

This notation of complex number gives as a very convenient tool for writing the sum, product etc. of the complex numbers. One can now add, multiply etc. the complex numbers of the types $a+ib$ in the same way as we do for real numbers, remembering that $i^{2}=-1$.
If $z = a+ib$, $a, b \in R$ is a complex number, then $a$ is called the Real part of the complex number, denoted by $Re(z)$ and $b$ is called the Imaginary part of the complex numbers, denoted by Im(z).

The symbol $i$ is called iota, with the property that
$i^{2} = -1$
i.e., $i = \sqrt{-1}$
1. If a complex number $z = a+ib$ is equal to zero, then the real part is separately equal to zero and the imaginary part is also separately equal to zero.

i.e., If $z = a+ib = 0$

then, $a = 0$, $b = 0$.

2. If two complex numbers are equal, then their real parts are equal and also their imaginary parts are equal.

i.e., if $z_{1} = a+ib$ and $z^{2} = c+id$

If $z_{1} = z_{2}$

$\Rightarrow a+ib$ = $c+id$

then, $a =c$, $b = d$.
The basic operations of complex numbers are given below.

1. Addition of complex number
Let $z_{1} = a_{1}+ib_{1}$ and $z_{2} = a_{2}+ib_{2}$ be two complex numbers. Then, their sum $z_{1} + z_{2}$ is given as

$z_{1} + z_{2}$ = $(a_{1}+ib_{1}) + (a_{2}+ib_{2})$

= $(a_{1}+ a_{2}) + i(b_{1} + b_{2})$

2. Subtraction of complex number
Let $z_{1} = a_{1}+ib_{1}$ and $z_{2} = a_{2}+ib_{2}$ be two complex numbers. Then, their difference is denoted as $z_{1} - z_{2}$ and is defined as the addition of $z_{1}$ and $-z_{2}$.

i.e., $z_{1} – z_{2}$ = $z_{1} +(-z_{2})$

= $(a_{1}+ib_{1})$ + $(-a_{2}- ib_{2})$

= $(a_{1} - a_{2}) + i(b_{1} – b_{2})$

3. Multiplication of complex number
Let $z_{1} = a_{1}+ib_{1}$ and $z_{2} = a_{2}+ib_{2}$ be two complex numbers. Then, their multiplication is denoted by $z_{1}z_{2}$ and is given as

$z_{1} z_{2}$ = $(a_{1}+ib_{1})(a_{2}+ib_{2})$
= $(a_{1}a_{2}-b_{1}b_{2}) + i(a_{1}b_{2}+a_{2}b_{1})$

4. Division of complex number
The division of a complex number $z_{1}$ by a non-zero complex number $z_{2}$ is defined as the multiplication of $z_{1}$ by the multiplicative inverse of $z_{2}$ and is denoted by $\frac{z_{1}}{z_{2}}$.

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

Let $z_{1} = a+ib$ and $z_{2} = c+id$ be two complex numbers, where $z_{2} \neq 0$ and $a, b, c, d \in R$

$\frac{z_{1}}{z_{2}}$ = $\frac{a+ib}{c+id}$ = $\frac{ac+bd}{c^{2}+d^{2}}$+ $i$ $\frac{bc-ad}{c^{2}+d^{2}}$

Multiplicative inverse of $z = a+ib$ is equal to $\frac{1}{z}$ = $\frac{1}{a+ib}$
The following are the examples of complex numbers.

Solved Examples

Question 1: Find the sum of the complex numbers $6+3i$ and $3-6i$?
Solution:
 
Given $z_{1} = 6+3i$ and $z_{2} = 3-6i$

$z_{1} + z_{2}$ = (6+3) + i(3-6)
 
= $9 -3i$

 

Question 2: Find the difference between the complex numbers $4+3i$ and $3+2i$?
Solution:
 
Given $z_{1} = 4+3i$ and $z_{2} = 3+2i$

$z_{1} - z_{2}$ = $z_{1} + (- z_{2})$ = $4+3i +(-3-2i)$

= $1+ i$

 

Question 3: Find the product of complex numbers $4+3i$ and $3+2i$?
Solution:
 
Given $z_{1} = 4+3i$ and $z_{2} = 3+2i$

$z_{1}z_{2}$ =  $(4+3i) (3+2i)$

= $(4 \times 3 – 3 \times 2) + i(4 \times 2 + 3 \times 3)$

= $(12 – 6) + i(8 + 9)$

= $6 + 17i$
 

Question 4: Simplify $\frac{5+4i}{4+5i}$
Solution:
 
$\frac{5+4i}{4+5i}$ = $\frac{5+4i}{4+5i}$ $\times$ $\frac{4-5i}{4-5i}$ 

= $\frac{(5+4i)(4-5i)}{(4+5i)(4-5i)}$

= $\frac{20 + 16i – 25i -20i^{2}}{4^{2}-5^{2}}$

= $\frac{40-9i}{16+25}$ = $\frac{40}{41}$-$\frac{9}{41}$ $i$