In this unit you are going to learn about the **modulus **and **argument **of a complex number. These

are quantities which can be recognised by looking at an Argand diagram. Recall that any complex

number, z, can be represented by a point in the complex plane as shown in Figure 1.

The complex number z is represented by point P . Its modulus and argument are shown.

We can join point P to the origin with a line segment, as shown. We associate with this line segment

two important quantities. The length of the line segment, that is OP , is called the **modulus **of the

complex number. The angle from the positive axis to the line segment is called the **argument **of

the complex number, z.

The modulus and argument are fairly simple to calculate using trigonometry.

**Example. **Find the modulus and argument of z = 4 + 3i.

**Solution. **The complex number z = 4 + 3i is shown in Figure 2. It has been represented by the

point Q which has coordinates (4, 3). The modulus of z is the length of the line OQ which we can

find using Pythagoras’ theorem.

(OQ)2 = 42 + 32 = 16 + 9 = 25

and hence OQ = 5.

Hence the modulus of z = 4 + 3i is 5. To find the argument we must calculate the angle between

the x axis and the line segment OQ. We have labelled this θ in Figure 2.

By referring to the right-angled triangle OQN in Figure 2 we see that

tan θ = 3/4

θ = tan−1 3/4 = 36.97◦

To summarise, the modulus of z = 4 + 3i is 5 and its argument is θ = 36.97◦. There is a special

symbol for the modulus of z; this is |z|. So, in this example, |z| = 5. We also have an abbreviation

for argument: we write arg(z) = 36.97◦.

When the complex number lies in the first quadrant, calculation of the modulus and argument is

straightforward. For complex numbers outside the first quadrant we need to be a little bit more

careful. Consider the following example.

**Example.**

Find the modulus and argument of z = 3 − 2i.

**Solution. **The Argand diagram is shown in Figure 3. The point P with coordinates (3, −2) represents

z = 3 − 2i.

We use Pythagoras’ theorem in triangle ONP to find the modulus of z:

(OP )2 = 32 + 22 = 13

OP = √13

Using the symbol for modulus, we see that in this example |z| = √13.

We must be more careful with the argument. When the angle θ shown in Figure 3 is measured in a

clockwise sense convention dictates that the angle is negative. We can find the size of the angle by

referring to the right-angled triangle shown. In that triangle tan α = 2 3 so that α = tan−1 2 3 = 33.67◦.

This is not the argument of z. The argument of z is θ = −33.67◦. We often write this as

arg(z) = −33.67◦.

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