# What are Complex Numbers? Operations on Complex Numbers

## What are Complex Numbers?

As we already know that the square of a real number is positive, therefore, the solution of the below-mentioned equation is not available in Real numbers. $\displaystyle {{x}^{2}}+1=0\,\,\Rightarrow \,\,{{x}^{2}}=-1$

Therefore, in order to remove this deficiency of real numbers, we require a number whose square must be -1.

Hence, the mathematicians introduced a larger set of number which is called the set of complex numbers that includes real numbers and every number whose square is not positive.

They designed another number i.e. – 1 which is called an imaginary unit. Complex Number is denoted by the a letter $\displaystyle i$ which contain the property that $\displaystyle {{i}^{2}}=-1$.

Leonard Euler was a Swiss Mathematician who first used the symbol $\displaystyle i$ for the number $\displaystyle \sqrt{{-1}}$.

It is pertinent to mention here that, the $\displaystyle i$ is not a real number.

Complex numbers are represented by Z and the set of all complex numbers is represented by C.

## Pure Imaginary Number

It is a square root of a negative real number. For example, $\displaystyle \sqrt{{-1}},\,\sqrt{{-11}},\,\,\sqrt{{-23}}$.

## Basic Arithmetic Operations on Complex Numbers

Consider, $\displaystyle {{Z}_{1}}=a+ib\,\,\,and\,\,\,\,{{Z}_{2}}=c+id$ are two complex numbers, where a, b, c and d belong to the real numbers then the sum of these two complex numbers will be $\displaystyle {{Z}_{1}}+{{Z}_{2}}=a+ib\,+c+id=a+c+i(b+d)$

Examples:

a – $\displaystyle a.\,(4-i)+(12-20i)$

Solution: $\displaystyle 4-i+12-20i=16-21i$

b – $\displaystyle (14-i)+(10-6i)$

Solution: $\displaystyle 14-i+10-6i=24-7i$

c – $\displaystyle (2-i)+(8-3i)$

Solution: $\displaystyle 2-i+8-3i=10-4i$

### 2. Subtraction

Consider, $\displaystyle {{Z}_{1}}=a+ib\,\,\,and\,\,\,\,{{Z}_{2}}=c+id$ are two complex numbers, where a, b, c and d belong to the real numbers then the difference between these two complex numbers will be $\displaystyle {{Z}_{1}}-{{Z}_{2}}=(a+ib\,)-(c+id)=a+ib-c-id=a-c+i(b-d)$

Examples:

a – $\displaystyle (4-i)-(12-20i)$

Solution: $\displaystyle 4-i-12+20i=-8+19i$

b – $\displaystyle (14-i)-(10-6i)$

Solution: $\displaystyle 14-i-10+6i=4+5i$

c – $\displaystyle (2-i)-(8-3i)$

Solution: $\displaystyle 2-i-8+3i=-6+2i$

### 3. Multiplication:

Consider, $\displaystyle {{Z}_{1}}=a+ib\,\,\,and\,\,\,\,{{Z}_{2}}=c+id$ are two complex numbers, where a, b, c and d belong to the real numbers then the product of these two complex numbers will be $\displaystyle \begin{array}{l}{{Z}_{1}}{{Z}_{2}}=(a+ib\,)(c+id)=ac+aid+ibc+{{i}^{2}}bd\\=ac+aid+ibc+(-1)bd=ac+aid+ibc-bd=(ac-bd)+(ad+bc)i\end{array}$

Some multiplication examples of real numbers are given here.

1. $\displaystyle \begin{array}{l}=16-6i-8i+3{{i}^{2}}=16-14i+3(-1)\\=16-14i-3=13-14i\end{array}$
2. $\displaystyle \begin{array}{l}(3-2i)(4+5i)\\=12+15i-8i-10{{i}^{2}}=12+7i-10(-1)\\=12+10+7i=22+7i\end{array}$

### 4. Division

Consider, $\displaystyle {{Z}_{1}}=a+ib\,\,\,and\,\,\,\,{{Z}_{2}}=c+id$ are two complex numbers, such that, $\displaystyle {{Z}_{2}}\ne 0$, the division of these complex numbers is $\displaystyle \frac{{{{Z}_{1}}}}{{{{Z}_{2}}}}=\frac{{a+bi}}{{c+di}}$

Few division examples of real numbers are given below: –