Modulus of Complex Number

Modulus of Complex Number

Modulus of the complex number is the distance of the point on the argand plane representing the complex number z from the origin.

Let P is the point that denotes the complex number z = x + iy.

Then distance OP = |z| = √(x2 + y2 ).

Note:

1. |z| > 0.

2. All the complex number with same modulus lie on the circle with centre origin and radius r = |z|.

Properties of Modulus of Complex Number

Some important properties of modulus of complex number and their proofs.

(i) |z1 z2| = |z1||z2|

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

Then, |z1 z2| = |(a + ib)(c + id)|

|ac + iad + ibc + i2bd|

|ac + iad + ibc – bd|

|ac – bd + i(ad + bc)|

(ac – bd)2+ (ad+bc)2

(ac)2+ (bd)2 – 2abcd + (ad)2 + (bc)2 + 2abcd

a2 c2 + b2 d2 + a2 d2+ b2 c2

a2 c2 + b2 c2 + b2 d2 + a2 d2

(a2 + b2)c2 + (b2 + a2)d2

(a2 + b2) (c2 + d2)

|z1||z2|.

(ii) |z/ z2 | = (|z1|) / (|z2|).

Proof: |z1/z2 | = |z1 . 1/z2 |

Using multiplicative property of modulus we have

|z1| |1/z2|

|z1| 1/(|z2|)

(|z1 |) / (|z2|).

Some other important results:

Triangle inequalities:

|z1 + z2| ≤ |z1| + |z2|

|z1 + z2| ≥ |z1| – |z2|

|z1 – z2| ≥ |z1| – |z2|.

Examples on Modulus of a Complex Number

Example : Find the modulus of the complex number z = (3 – 2i)/2i

Solution: z = (3 – 2i)/2i

z = (3 )/2i – 2i/2i

z = 3/2i – 1

z = 3i/(2i2 ) – 1

z = (-3i/2)-1