# Conjugate of a Complex Number

## What is a Complex Number?

### A number that can be represented in the form of (a + ib), where ‘i’ is an imaginary number called iota, can be called a complex number. A complex number is basically a combination of a real part and an imaginary part of that number.

For example, 6 + i3 is a complex number in which 6 is the real part of the number and i3 is the imaginary part of the number.

What Is a Conjugate?

A conjugate in Mathematics is formed by changing the sign of one of the terms in a binomial. For example, if the binomial number is a + b, so the conjugate of this number will be formed by changing the sign of either of the terms. If we change the sign of b, so the conjugate formed will be a – b. Therefore, in mathematics, a + b and a – b are both conjugates of each other.

Conjugate of a Complex Number?

The conjugate of a complex number represents the reflection of that complex number about the real axis on Argand’s plane. When the i of a complex number is replaced with -i, we get the conjugate of that complex number that shows the image of that particular complex number about the Argand’s plane.

For example, as shown in the image on the right side, z = x + iy is a complex number that is inclined on the real axis making an angle of α and z = x – iy which is inclined to the real axis making an angle -α.

### Properties of the conjugate of a Complex Number

1. $\overline{z_{1} \pm z_{2} }$ = $\overline{z_{1}}$  $\pm$ $\overline{z_{2}}$

Proof: Let z1 = p + iq and z2 = x + iy

So, $\overline{z_{1} \pm z_{2} }$ = $\overline{p + iq \pm + iy}$

= $\overline{p \pm x + iq \pm iy}$

= $\overline{p \pm x – i(q \pm y)}$

= $\overline{p – iq \pm x – iy}$

=  $\overline{z_{1}}$ $\pm$ $\overline{z_{2}}$

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

Proof: Let z1 = a + ib and z2 = c + id

Then, $\overline{z_{}. z_{2}}$ =  $\overline{(a + ib) . (c + id)}$

= $\overline{ac + iad + ibc + i2bd}$

= $\overline{ac + iad + ibc – bd}$

= $\overline{ac – bd + i(ad + bc)}$

= ac -bd – i(ad + bc)

= ac + i2bd – iad – ibc

= c(a – ib) – id(a – ib)

= (a – ib).(c – id)

= $\overline{z_{}. z_{2}}$

3. $\frac{\overline{z_{1}}}{z_{2}}$ =  $\frac{\overline{z}_{1}}{\overline{z}_{2}}$

Proof, $\frac{\overline{z_{1}}}{z_{2}}$ =    $\overline{(z_{1}.\frac{1}{z_{2}})}$

Using the multiplicative property of conjugate, we have

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

$\frac{\overline{z}_{1}}{\overline{z}_{2}}$

4. $\overline{z}$ = z

Proof: Let z= a + ib

Then, $\overline{z}$ =  $\overline{a + ib}$ = $\overline{a – ib}$ = a + ib = z

5 .If z = a + ib

Then, z. $\overline{z}$  = a2 + b2 = |z2|

Proof: z.  $\overline{z}$ = (a + ib). $\overline{(a + ib)}$ = (a + ib).(a – ib) = a2 – i2b2 = a2 + b2 = |z2|

6.  z +  $\overline{z}$ = x + iy + ( x – iy )

## Then,  z + $\overline{z}$ = 2x

7.  z –  $\overline{z}$ = x + iy – ( x – iy )

Then, z – $\overline{z}$ = 2 iy

Solved Problems

Question 1. Identify the conjugate of the complex number 5 + 6i. Describe the real and the imaginary numbers separately.

The conjugate of the complex number 5 + 6i  is 5 – 6i.

The real part of the resultant number = 5 and the imaginary part of the resultant number = 6i.

Question 2. Find all the complex numbers of the form z = p + qi , where p and q are real numbers such that z. $\overline{z}$ = 25 and p + q = 7 where $\overline{z}$ is the complex conjugate of z.

Answer: It is given that z. $\overline{z}$ = 25

By the definition of the conjugate of a complex number,

$\overline{z}$ = p – iq

Therefore, z. $\overline{z}$  = (p + iq) . (p – iq) = 25

= p2 + q2 = 25

Since, p + q = 7 , q = 7 – p

Therefore, p2 + q2 = p2 + ( 7- p )2 = 25

= p2 + 49 + p2 – 14 p = 25

= 2p2 – 14p + 24 = 0

= p2 – 7p + 12 = 0

= p(p-3) – 4(p-3) = 0

= (p – 3).(p – 4) =0

Therefore, either p = 3 or p =4

So, if p=3, q=4 ; else if p =4, q = 3

Possible complex numbers are: 3 + i4 or 4 + i3.