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.