# How to Find the Modulus of a Complex Number

Here you will learn how to find the modulus of a complex number and properties of modulus with examples.

Let’s begin –

## How to Find the Modulus of a Complex Number

The modulus of a complex number z = a + ib is denoted by | z | and is defined as

| z | = $$\sqrt{a^2 + b^2}$$ = $$\sqrt{[Re (z)]^2 + [Im (z)]^2}$$

Clearly, | z | $$\ge$$ 0 for all z $$\in$$ C.

Example : If $$z_1$$ = 3 – 4i, $$z_2$$ = -5 + 2i and $$z_3$$ = 1 + $$\sqrt{-3}$$, then find modulus of $$z_1$$, $$z_2$$ and $$z_3$$.

Solution : We have, $$z_1$$ = 3 – 4i, $$z_2$$ = -5 + 2i and $$z_3$$ = 1 + $$\sqrt{-3}$$

| $$z_1$$ | = | 3 – 4i | = $$\sqrt{3^2 + (-4)^2}$$ = 5,

| $$z_2$$ | = | 5 + 2i | = $$\sqrt{(-5)^2 + 2^2}$$ = $$\sqrt{29}$$

and, | $$z_3$$ | = | 1 + $$\sqrt{-3}$$ | = $$\sqrt{1^2 + (\sqrt{3})^2}$$ = 2

Remark : In the set C of all complex numbers, the order relation is not defined. As such $$z_1$$ > $$z_2$$ or $$z_1$$ < $$z_2$$ has no meaning but | $$z_1$$ | > | $$z_2$$ | or | $$z_1$$ | < | $$z_2$$ | has got its meaning since | $$z_1$$ | and | $$z_2$$ | are real numbers.

### Properties of Modulus

If z, $$z_1$$, $$z_2$$ $$\in$$ C, then

(i) | z | = 0 $$\iff$$ z = 0  i.e.  Re (z) = Im (z) = 0

(ii) | z | = | $$\bar{z}$$ | = | -z |

(iii) – | z | $$\le$$ Re (z) $$\le$$ | z |  ;  – | z | $$\le$$ Im (z) $$\le$$ | z |

(iv) $$z\bar{z}$$ = $$| z |^2$$

(v) | $$z_1 z_2$$ | = | $$z_1$$ | | $$z_2$$ |

(vi) | $$z_1\over z_2$$ | = $$| z_1 | \over | z_2 |$$ ,  $$z_2$$ $$\ne$$ 0

(vii)  $$| z_1 + z_2 |^2$$ = $$| z_1 |^2$$ + $$| z_2 |^2$$ + $$2 Re (z_1\bar{z_2})$$

(viii)  $$| z_1 – z_2 |^2$$ = $$| z_1 |^2$$ + $$| z_2 |^2$$ – $$2 Re (z_1\bar{z_2})$$

(ix)  $$| z_1 + z_2 |^2$$ + $$| z_1 – z_2 |^2$$ = 2($$| z_1 |^2$$ + $$| z_2 |^2$$)

(x) $$| az_1 – bz_2 |^2$$ + $$| bz_1 + az_2 |^2$$ = $$a^2 + b^2$$ ($$| z_1 |^2$$ + $$| z_2 |^2$$), where a. b $$\in$$ R.