# 2 Complex numbers

## 2.8 Polar form

You have seen that the complex number x + iy corresponds to the point (x, y) in the complex plane. This correspondence enables us to give an alternative description of complex numbers, using so-called polar form. This form is particularly useful when we discuss properties related to multiplication and division of complex numbers.

Polar form is obtained by noting that the point in the complex plane associated with the non-zero complex number z = x + iy is uniquely determined by the modulus ,

together with the angle θ (measured in an anticlockwise direction in radians) between the positive direction of the x-axis and the line from the origin to the point, as shown in the graph above. We have

so the complex number z can be expressed as

This description of z in terms of r and θ is not unique because the angle θ is determined only up to multiples of 2; that is, the angles θ ± 2, θ ± 4, θ ± 6, …, also determine the same complex number. However, if we restrict the angle θ to lie in the interval (−, ], then the description is unique. The origin is an exception, however: at the origin the value of r is 0, and θ is not defined.

Some texts restrict θ to lie in the interval [0, 2).

### Definitions

A non-zero complex number z = x + iy is in polar form if it is expressed as

where r = |z| and θ is any angle (measured in radians anticlockwise) between the positive direction of the x-axis and the line joining z to the origin.

(Some texts use r cis θ or 〈r, θ 〉 as shorthand for r(cos θ + i sin θ).)

Such an angle θ is called an argument of the complex number z, and is denoted by arg z. The principal argument of z is the value of arg z that lies in the interval (−, ], and is denoted by Arg z.

Here principal argument is a shortened form of the more conventional ‘principal value of the argument’.

Remark

Sometimes we refer to z = x + iy as the Cartesian form of z, to distinguish it from polar form.

We now discuss how to convert a complex number from polar form to Cartesian form, and vice versa.

When carrying out such conversions, it is useful to remember the values in the table below, as these will help you in some special cases. You may find it easier to remember the triangles, from which you can work out most of the values in the table.

The following formulas are also helpful. For any θ :

You will need only the first two equations in each row; the other equations are given for completeness.

You may be able to remember these formulas by roughly sketching graphs of the sine and cosine functions, and using their symmetry. For example, we can sketch the sine function as follows.

To convert a complex number from polar form to Cartesian form is straightforward: we use the equations

to find x and y given r and θ.

This is illustrated in the following example.

### Example 3

Express each of the following complex numbers in Cartesian form.

(a)  3(cos(/3) + i sin(/3))
(b)  cos(−/6) + i sin(−/6)

### Solution

(a)  Here r = 3 and θ = /3. Thus the required form is x + iy, where

and

The Cartesian form is therefore
.
(b)  Here r = 1 and θ = −/6. Thus the required form is x + iy, where

and

The Cartesian form is therefore
.

### Exercise 15

Express each of the following complex numbers in Cartesian form.

(a)  2(cos(/2) + i sin(/2))
(b)  4(cos(−2/3) + i sin(−2/3))

### Solution

(a)  The required form is x + iy, where

and

The Cartesian form is therefore 2i.
(b)  The required form is x + iy, where

and

The Cartesian form is therefore − 2 − 2i.

To convert a non-zero complex number z from Cartesian form x + iy to polar form r(cos θ + i sin θ), we first find the modulus r using the formula

If z is either real or imaginary, then it lies on one of the axes and has principal argument 0, /2, or −/2, as shown in diagram (a) below. Otherwise, to find the principal argument θ, we need to solve the equations

We can do this by first finding the first-quadrant angle φ that satisfies the related equation

The relationship of φ to the principal argument θ depends on the quadrant in which z lies, as indicated in diagram (b) below.

Note: φ is the angle at the origin in the right-angled triangle formed by drawing the perpendicular from z to the real axis, as illustrated above, in the case where z lies in the second quadrant. We have

and the relationship between θ and φ can be seen to be as given in diagram (b).

The quadrant in which z lies can be found from the values of x and y, and θ can then be deduced by using the appropriate equation from diagram (b). (You may find it helpful to sketch z on an Argand diagram.)

This method for finding θ is used in the next example, which illustrates how the Cartesian form of a complex number is converted into polar form.

### Example 4

Express each of the following complex numbers in polar form, using the principal argument.

(a)  2 + 2i
(b)

### Solution

(a)  Let z = x + iy = 2 + 2i, so x = 2 and y = 2.
Then z = r(cos θ + i sin θ), where

To find θ, we calculate

So φ = /4, and z lies in the first quadrant so θ = φ = /4.
The polar form of 2 + 2i in terms of the principal argument is therefore

(b)  Let , so and .
Then z = r(cos θ + i sin θ), where

To find θ, we calculate

So φ = /3, and z lies in the third quadrant so θ = −(φ) = −2/3.
The polar form of in terms of the principal argument is therefore

### Exercise 16

Draw a diagram showing each of the following complex numbers in the complex plane, and express them in polar form, using the principal argument.

(a)  −1 + i
(b)  1 − i
(c)  −5

### Solution

(a)  Let z = x + iy = −1 + i, so x = −1 and y = 1. Then z = r(cos θ + i sin θ), where

Also

So φ = /4, and z lies in the second quadrant, so θ = φ = 3/4.
Thus the polar form of −1 + i in terms of the principal argument is

(b)  Let z = x + iy = 1 − i, so x = 1 and y = − . Then z = r(cos θ + i sin θ), where

Also

So φ = /3, and z lies in the fourth quadrant, so θ = − φ = − /3.
Thus the polar form of 1 − √3i in terms of the principal argument is

(c)  Let z = x + iy = −5, so x = −5 and y = 0. Then z = r(cos θ + i sin θ), where

Also z lies on the negative half of the real axis, so θ = .
Thus the polar form of −5 in terms of the principal argument is

Polar form gives us a simple way to multiply complex numbers. Let

then

(Here we use the addition formulas for the trigonometric functions.)

That is, to multiply two complex numbers in polar form, we multiply their moduli and add their arguments. For example,

Note:

In the rest of this section we usually write /4 as , and similarly for other fractions of . You may use either form, or the form , as you wish.

We can also use formula (2.1) for the product of two complex numbers in polar form to establish a similar formula for the quotient of two complex numbers. Specifically, we show that if

with z2 ≠ 0, then z1/z2 is the complex number

z2 ≠ 0 r2 ≠ 0

To see this, notice that since r1 = rr2 and θ1 = θ + θ2 it follows from the above discussion that z1 = zz2. Hence z1/z2 = z, as required. We can write the formula as

That is, to divide a complex number z1 by another complex number z2, we divide the modulus of z1 by the modulus of z2, and subtract the argument of z2 from the argument of z1.

For example,

In particular, if z = r(cos θ + i sin θ) with r ≠ 0, then the reciprocal of z is

The above methods for multiplying and dividing complex numbers in polar form are summarised below.

### Strategy 1

To multiply two complex numbers given in polar form, multiply their moduli and add their arguments.

To divide a complex number z1 by a non-zero complex number z2 when both are given in polar form, divide the modulus of z1 by the modulus of z2, and subtract the argument of z2 from the argument of z1.

Remark

If you require the principal argument of the product or quotient, then you may need to add or subtract 2 from the argument calculated.

### Exercise 17

Determine each of the following products and quotients in polar form in terms of the principal argument.

(a)
(b)
(c)
(d)

### Solution

(a)  The modulus of the product is

An argument is

Since this argument lies in (− , ], it follows that it is the principal argument. The required product is therefore

(b)  The modulus of the product is

An argument is

The principal argument is therefore

The required product is therefore

(c)  The modulus of the quotient is

An argument is

The principal argument is therefore

The required quotient is therefore

(d)  The modulus of the quotient is

An argument is

Since this argument lies in (− , ], it follows that it is the principal argument. The required quotient is therefore