# What are Complex Numbers?

• Last Updated : 29 Mar, 2022

A complex number is a term that can be shown as the sum of real and imaginary numbers. These are the numbers that can be written in the form of a + ib, where a and b both are real numbers. It is denoted by z. Here the value ‘a’ is called the real part which is denoted by Re(z), and ‘b’ is called the imaginary part Im(z) in form of a complex number.  It is also called an imaginary number. In complex number form a + bi ‘i’ is an imaginary number called “iota”. The value of i is (√-1) or we can write as i2 = -1. For example,

9 + 16i is a complex number, where 9 is a real number (Re) and 16i is an imaginary number (Im).

10 + 20i is a complex number where 10 is a real number (Re) and  20i is an imaginary number (Im)

### Algebraic Operations on Complex number

A real number and imaginary number combination are called a Complex number. There are four types of algebraic operation of complex numbers,

In this operation, we know that a complex number is of the form z = p + iq where a and b are real numbers. Now, consider two complex numbers z1 = p1 + iq1 and z2 = p2 + iq2. Therefore,  the addition of the complex numbers z1 and z2.

z1 + z2 = (p1 + p2) + i(q1 + q2)

Some more identities are:

z1 + z2 = z

z1 + z2 = z2 + z1

(z1 + z2) + z3 = z1 + (z2 + z3)

z + (-z) = 0

(p + iq) + (0 + i0) = p + iq

The resulting complex number real part is the sum of the real part of each complex number. The resulting complex number imaginary part is equal to the sum of the imaginary part of each complex number.

Subtraction of Complex Numbers

In this operation of the complex numbers z1 = p1 + iq1 and z2 = p2 + iq2, therefore the difference of z1 and z2  which is z1-z2 is defined as,

z1 – z2 = (p1 – p2) + i(q1 – q2)

Multiplication of Complex Numbers

In this operation of multiplication of Two Complex Numbers. We know that (x + y)(z + w).

= xz + xw + zy + zw

Similarly, the complex numbers z1 = p1 + iq1 and z2 = p2 + iq2

To find z1z2:

z1 z2 = (p1 + iq1)(p2 + iq2)

z1 z2 = p1 p2 + p1 q2i + q1 p2i + q1q2i2

As we know,  i2 = -1,

Therefore,

z1 z2 = (p1 p2 – q1 q2) + i(p1 q2 + p2 q1)

Some identities are:

z1 × z2 = z

z1.z2 = z2.z1

z1(z2.z3) = (z1.z2)z3

z1(z2 + z3) = z1.z2 + z1.z3

Division of Complex Numbers

In this operation of complex number z1 = p1 + iq1 and z2 = p2 + iq2, therefore, to find z1/z2, we have to multiply the numerator and denominator with the conjugate of z2.

The division of complex numbers:

Let z1 = p1 + iq1 and z2 = p2 + iq2,

z1/z2 = (p1 + iq1)/(p2 + iq2)

Hence, (p1 + iq1)/(p2 + iq2) = [(p1 + iq1)(p2 – iq2)] / [(p2 + iq2)(p2 – iq2)]

(p1 + iq1)/(p2 + iq2) = [(p1p2) – (p1q2i) + (p2q1i) + q1q2)] / [(p22 + q22)]

(p1 + iq1)/(p2 + iq2) = [(p1p2) + (q1q2) + i(p2q1 – p1q2)] / (p22 + q22)

z1/z2 = (p1p2) + (q1q2) / (p22 + q22) + i(p2q1 – p1q2) / (p22 + q22)

Rules of imaginary numbers

i = √-1

i2 = -1

i3 = -i

i4 = 1

i4n = 1

i4n-1 = -1

### Sample Questions

Question 1: Perform the indicated operation and write the answer in standard form: 4+5i/4-5i

Solution:

Given : 2+5i/2-5i

to simplifying multiply the numerator and denominator by the conjugate of denominator

= (2+5i/2-5i) × (2+5i)(2+5i)

= {(2+5i)2}/ {(2)2– (5i)2}

= {4 + (5i)2 + 2(2)(5i)} / {4 – 25(i)2}

= {4 +25(i)2 + 20i} / {4 +25}

= {4 – 25 + 20i} / 29

= (-21 + 20i) / 29

= -21/29 + 20/29i

Question 2: Simplify (5 + 2i) – (9 + 7i), Find the square of imaginary number?

Solution:

Given : (5 + 2i) – (9 + 7i)

= 5+2i -9 -7i

= (5 – 9) + (2 – 7)i

= (-4 – 5i)

Now imaginary number = -5i

(-5i)2 = -5i x -5i

= 25i2

= 25(-1)

= – 25

Question 3: Find the additive inverse of complex number 9 + 7i?

Solution:

It is defined as the value which on adding with the original number results in  zero value. An additive inverse of a complex number is the value we add to a number to yield zero.

So here the additive inverse of complex number 9 + 7i is -(9 + 7i)

= -9 – 7i

Question 4: Express (5 -.3i)/(9 + 2i) in standard form?

Solution:

Given: (5 – 3i)/(9 + 2i)

Multiplying with the conjugate of denominators,

= {(5 – 3i)/(9 + 2i) × (9 – 2i)/(9 – 2i)}

= {(5 – 3i)(9 – 2i)} / {(9)2 – (2i)2}

= {45- 10i – 27i + 6i2} / (81- 4i2)

= {45 – 37i + 6(-1)} / (81+4)

= ( 39 – 37i) / 85

= 39/85 – 37/85i

Question 5: Find the multiplicative inverse of 2 – 8i?

Solution:

The multiplicative inverse of a complex number z is simply 1/z.

It is denoted as : 1 / z  or  z-1 (Inverse of z)

here z = 2 – 8i

Therefore z = 1/z

= 1 / (2 – 8i)

Now rationalizing

= 1/(2 – 8i) x (2 + 8i)/(2 +8i)

= (2 + 8i) / {(2)2 – 82i2}

= (2 + 8i) / { 4 + 64}

= (2 + 8i)/ (68)

= 2/68 + 8i/68

= 1/34 + 2/17 i

Question 6: Simplify: (9-4i)(7-7i).

Solution:

Given:  (9-4i)(7-7i)

= 63 – 63i -28i +28i2

= 63 – 63i – 28i + 28(-1)

= 63- 91i – 28

= 35 – 91i

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