## Complex Number Polar Form Argument Essay

## 4. Polar Form of a Complex Number

by M. Bourne

We can think of complex numbers as **vectors**, as in our earlier example. [See more on Vectors in 2-Dimensions].

We have met a similar concept to "polar form" before, in Polar Coordinates, part of the analytical geometry section.

### Why Polar Form?

In the Basic Operations section, we saw how to add, subtract, multiply and divide complex numbers from scratch.

However, it's normally much easier to multiply and divide complex numbers if they are in **polar form**.

Our aim in this section is to write complex numbers in terms of a distance from the origin and a direction (or angle) from the positive horizontal axis.

We find the real (horizontal) and imaginary (vertical) components in terms of *r* (the length of the vector) and *θ* (the angle made with the real axis):

From Pythagoras, we have: `r^2=x^2+y^2` and basic trigonometry gives us:

`tan\ theta=y/x``x=r\ cos theta``y = r\ sin theta`

Multiplying the last expression throughout by `j` gives us:

`yj = jr\ sin θ`

So we can write the **polar form** of a complex number as:

`x + yj = r(cos θ + j\ sin θ)`

*r* is the **absolute value** (or **modulus**) of the complex number

*θ* is the **argument** of the complex number.

There are two other ways of writing the **polar form** of a complex number**:**

`r\ "cis"\ θ` [This is just a shorthand for `r(cos θ + j\ sin θ)`]

`r\ ∠\ θ` [means once again, `r(cos θ + j\ sin θ)`]

NOTE: When writing a complex number in polar form, the angle *θ* can be in DEGREES or RADIANS.

### Example 1

Find the polar form and represent graphically the complex number `7 - 5j`.

Answer

### Example 2

Express `3(cos 232^@+ j sin 232^@)` in rectangular form.

Answer

### Exercises

**1. **Represent `1+jsqrt3` graphically and write it in polar form.

Answer

**2. **Represent `sqrt2 - j sqrt2` graphically and write it in polar form.

Answer

**3. **Represent graphically and give the rectangular form of `6(cos 180^@+ j\ sin 180^@)`.

Answer

**4. **Represent graphically and give the rectangular form of `7.32 ∠ -270°`

Answer

## And the good news is...

Now that you know what it all means, you can use your calculator directly to convert from **rectangular to polar** forms and in the other direction, too.

How to convert polar to rectangular using hand-held calculator.

### Online Polar Calculator

Also, don't miss this interactive polar converter graph, which converts from polar to rectangular forms and vice-versa, and helps you to understand this concept:

Online polar to rectangular calculator

## 1 Opening items

## 1.1 Module introduction

The symbol *i* is defined to have the property that *i* × *i* = −1. Expressions involving *i*, such as 3 + 2*i*, are known as *complex numbers*, and they are used extensively to simplify the mathematical treatment of many branches of physics, such as oscillations, waves, a.c. circuits, optics and quantum theory. This module is concerned with the representation of *complex numbers* in terms of *polar coordinates*, together with the related *exponential* representation. Both representations are particularly useful when considering the multiplication and division of complex numbers, and are widely used in physics.

In Subsection 2.1 we review the *Cartesian representation* of complex numbers and show how any complex number can be represented as a point on an *Argand diagram* (the *complex plane*). We also show how complex numbers can be interpreted as an *ordered pair* of real numbers. Points in a plane are often specified in terms of their *Cartesian coordinates*, *x* and *y*, but they can equally well be defined in terms of polar coordinates *r* and *θ*. It will transpire that, while addition and subtraction of complex numbers is easy for complex numbers in Cartesian form, multiplication and division are usually simplest when the numbers are expressed in terms of polar coordinates. Subsection 2.2 and the subsequent two subsections are concerned with the polar representation of complex numbers, that is, complex numbers in the form *r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*). Subsection 2.5 introduces the *exponential representation*, *r*_{ }e^{iθ}. Section 3 is devoted to developing the arithmetic of complex numbers and the final subsection gives some applications of the polar and exponential representations which are particularly relevant to physics.

## 1.2 Fast track questions

**Question F1**

The complex number *z* is defined by *z* = 1 + *i*. Find the following in their simplest representations: *z*, |_{ }*z*_{ }|, arg(*z*), *z** and *z*^{−1}.

Answer F1

We can write *z* as

$z = 1 + i = \sqrt{2\os}\left(\dfrac{1}{\sqrt{2\os}}+i\dfrac{1}{\sqrt{2\os}}\right)$

But we also know that

$\cos(\pi/4) = \dfrac{1}{\sqrt{2\os}}$ and $\sin(\pi/4) = \dfrac{1}{\sqrt{2\os}}$

and therefore

*z* = 2_{ }[cos(*π*/4) + *i*_{ }sin(*π*/4)]

which is the polar representation of *z*.

The modulus of *z* is given by |_{ }*z*_{ }| = $\sqrt{2\os}$. The argument of *z* is given by arg(*z*) = *π*/4. (The argument is undefined up to an additive term of 2*πn*, where *n* is any integer.)

The complex conjugate of *z* is given by

$z\cc = \sqrt{2\os}[\cos(\pi/4) - i\sin(\pi/4)]$

The reciprocal of *z* is given by

$z^{-1}= \dfrac{1}{\sqrt{2\os}}[\cos(\pi/4) - i\sin(\pi/4)]$

because $z^{-1} = \dfrac{z\cc}{\lvert\,z\,\rvert^2}$

**Question F2**

The complex numbers *z* and *w* are defined by *z* = 3_{ }e^{iπ/10} and *w* = 4_{ }e^{iπ/5}. Find the simplest exponential representations of *zw* and *z*/*w*.

Answer F2

*zw* = 3_{ }e^{iπ/10} × 4_{ }e^{iπ/5} = 3 × 4_{ }e^{i (π/10 + π/5)} = 12_{ }e^{3iπ/10}

$\dfrac{z}{w} = \dfrac{3{\rm e}^{i\pi/10}}{4{\rm e}^{i\pi/5}} = \dfrac{3}{4}{\rm e}^{i(\pi/10-\pi/5)} = \dfrac{3}{4}{\rm e}^{-i\pi/10}$

**Question F3**

Suppose that a complex quantity, *z*, is known to satisfy

*z* = 2 + *i* + e^{iθ}

where *θ* can take any real value. Sketch a curve on an Argand diagram giving the position of all possible points representing *z*.

**Figure 13** See Answer F3.

Answer F3

All complex numbers which satisfy *z* = e^{i}* ^{θ}* lie on a circle of radius 1, centre the origin, on an Argand diagram. The additional terms, 2 +

*i*, have the effect of moving the centre to the point representing 2 +

*i*, so the curve on which

*z*= 2 +

*i*+ e

^{i}

*must lie is shown in Figure 13.*

^{θ}## 1.3 Ready to study?

**Question R1**

Rationalize the expression $z = \dfrac{3 + 2i}{1+2i}$ (i.e. express *z* in the form *x* + *iy*, finding the values of the real numbers *x* and *y*). What are the real and imaginary parts of *z*? Also find the complex conjugate and the modulus of *z*.

Answer R1

To rationalize the expression we multiply the *denominator* and the *numerator* by the complex conjugate of the denominator

$z = \dfrac{3 + 2i}{1 + 2i} = \dfrac{3 + 2i}{1 + 2i} \times \dfrac{1 - 2i}{1 - 2i}$

$\phantom{z }= \dfrac{(3\times 1 + 2\times 2)+i\,(-2\times 3 + 2\times 1)}{1^2+2^2} = \dfrac{7-4i}{5}$

The *real part* of *z* (often written as Re(*z*)) is 7/5 and the *imaginary part* of *z* (often written as Im(*z*)) is −4/5. Notice that Im(*z*) is *not* equal to −4*i*/5.

The *complex conjugate* of *z* = *x* + *iy* (sometimes written as $\bar{z}$ but as *z** in *FLAP*) is equal to *x* − *iy* and so in this case we have

$z\cc = x - iy = \dfrac{7+4i}{5}$

The *modulus* of *z* = *x* + *iy* (usually written as |_{ }*z*_{ }|) is defined by |_{ }*z*_{ }| = $\sqrt{x^2+y^2}$ and therefore we have

$\lvert\,z\,\rvert = \sqrt{x^2+y^2} = \dfrac{\sqrt{7^2+4^2}}{5} = \dfrac{\sqrt{65\os}}{5} = \sqrt{13/5\os} \approx 1.612$

Note that the modulus of a complex number cannot be negative, i.e. |_{ }*z*_{ }| ≥ 0.

Consult the relevant terms in the *Glossary* for further information.

**Question R2**

Draw and label the points representing the complex numbers −2 + *i*, −2 − *i* and −3*i* on an Argand diagram.

**Figure 14** See Answer R2.

Answer R2

The points representing the complex numbers −2 + *i*, −2 − *i* and −3*i* are shown on the *Argand diagram* in Figure 14.

Consult *Argand diagram* in the *Glossary* for further information.

**Figure 1** See Question R3.

**Question R3 i**

The equilateral triangle shown in Figure 1 has a perpendicular drawn from one vertex to the opposite side. Use the triangle in Figure 1 to find the values of cos(*π*/3), sin(*π*/3), tan(*π*/3), cos(*π*/6), sin(*π*/6) and tan(*π*/6).

Answer R3

By *Pythagoras’s theorem* the length of the perpendicular in Figure 1 is $\sqrt{2^2-1^2} = \sqrt{3\os}$, so we have

cos(*π*/3) = 1/2, sin(*π*/3) = $\sqrt{3\os}$/2, tan(*π*/3) = $\sqrt{3\os}/1 = \sqrt{3\os}$

cos(*π*/6) = $\sqrt{3\os}$/2, sin(*π*/6) = 1/2, tan(*π*/6) = 1/$\sqrt{3\os}$

Consult *Pythagoras’s theorem* and *trigonometric functions* in the *Glossary* for further information.

**Figure 2** See Question R4.

**Question R4 i**

Use the right–angled triangle with two sides equal, shown in Figure 2 to find the values of cos(*π*/4), sin(*π*/4) and tan(*π*/4).

Answer R4

By *Pythagoras’s theorem* the length of the *hypotenuse* in Figure 2 is $\sqrt{1^2+1^2} = \sqrt{2\os}$ so we have

cos(*π*/4) = 1/$\sqrt{2\os}$, sin(*π*/4) = 1/$\sqrt{2\os}$, tan(*π*/4) = 1/1 = 1

Consult *inverse trigonometric functions* in the *Glossary* for further information.

**Question R5**

Two two–dimensional vectors, ** u** and

*, are specified in component form as (2, 3) and (1, −4), respectively. Find*

**v****+**

*u**by (a) drawing a suitable diagram and (b) adding the components directly.*

**v****Figure 15** See Answer R5.

Answer R5

(a) The two vectors are shown in Figure 15 and their sum is represented by the third side of the triangle.

(b) The sum of the vectors can also be obtained by adding their components,

(2, 3) + (1, −4) = (2 + 1, 3 −4) = (3, −1)

Consult *vector addition* in the *Glossary* for further information.

**Question R6**

Solve the equation tan_{ }*θ* = 1.

**Figure 16** See Answer R6.

Answer R6

From Answer R4 we know that *θ* = *π*/4 is one solution of the equation tan_{ }*θ* = 1.

However, from the graph of *y* = tan_{ }*θ* (see Figure 16) it is easy to see that *π*/4 ± *nπ* is also a solution for any integer *n*.

Consult *inverse trigonometric functions* in the *Glossary* for further information.

**Question R7**

Express $\left.\sqrt{{\rm e}^{3x}}\middle /{\rm e}^x\right.$ in the form e^{kx} (for some value of *k*) and hence write down the first three terms of its power series expansion.

Answer R7

$\left.\sqrt{{\rm e}^{3x}}\middle /{\rm e}^x\right. = {\rm e}^{3x/2} \times {\rm e}^{-x} = {\rm e}^{x/2} = 1 + x/2 + x^2/8 + \dots$

where the power series has been obtained by substituting *y* = *x*/2 into the general power series

${\rm e}^y = 1 + y + \dfrac{y^2}{2!} + \dfrac{y^3}{3!} + \dots$

Consult *power series* in the *Glossary* for further information.

## 2 Representing complex numbers

## 2.1 Complex numbers and Cartesian coordinates

A *complex number*, *z*, can be written as *z* = *x* + *iy* where *x* and *y* are real numbers and *i*^{2} = −1. Some examples of complex numbers are 2 + 3*i*, 7*i* and 2.4. i

Such numbers satisfy straightforward rules for addition and subtraction, which essentially mean that the real and imaginary parts are treated separately, so that, for example,

(3 + 4*i*) + (2 − *i*) − (−2 − 3*i*) = (3 + 2 + 2) + (4*i* − *i* + 3*i*) = 7 + 6*i*

**Figure 3** An Argand diagram showing the point corresponding to a complex number, *z* = *a* + *ib*.

**Figure 4** An Argand diagram showing the addition of 3 + 2*i* and 1 + 3*i*.

Multiplication is quite simple provided that we remember to replace every occurrence of *i* × *i* by −1, although a mathematician would probably prefer a formal statement that the product of two complex numbers (*a* + *ib*) and (*x* + *iy*) is given by

(*a* + *ib*)(*x* + *iy*) = (*ax* − *by*) + *i*_{ }(*ay* + *bx*)(7)

For a long time the meaning of the symbol *i* gave many famous mathematicians cause for concern. However, in 1833 Sir William Rowan Hamilton (1805–1865) realized that the *i* and + sign in *z* = *x* + *iy* are both unnecessary sources of confusion. The role of the *i* is really to keep the *x* and *y* separate, while the + sign is there to tell us that *x* and *y* are part of a single entity; it does not mean addition in the sense that we might, for example, add 2 apples to 3 apples to get 5 apples. In fact, the *x* and *y* in *z* = *x* + *iy* are very much like the (ordered) pairs of numbers used as Cartesian coordinates. Hamilton’s ideas are closely linked to the those of Robert Argand (1768–1822) and Karl Friedrich Gauss (1777–1855) who both suggested representing a complex number by a point in a plane. As an example, the complex number *a* + *ib* is shown on an (*x*, *y*) coordinate system in Figure 3; notice that it is conventional that the number multiplying the *i* corresponds to the *y*–value. A figure in which the real and imaginary parts of complex numbers are used as Cartesian coordinates is known as an *Argand diagram* or the *complex plane*. An expression such as *a* + *ib*, where *a* and *b* are real numbers, is said to be the **cartesian_form_of_a_complex_numberCartesian form** (or **cartesian_representation_of_a_complex_numberCartesian representation**) of a complex number.

In some mathematics textbooks the authors avoid the problem of the meaning of the symbol *i* entirely – by not mentioning it – and they introduce the complex numbers as a set of **ordered pairs** of real numbers (*x*, *y*) with certain operations defined on them. Such a treatment has the advantage that complex numbers can immediately be seen to have much in common with vectors. The addition of two complex numbers is then defined by

(*x*, *y*) + (*a*, *b*) = (*x* + *a*, *y* + *b*)(8)

which is just the same as the rule that defines the addition of two vectors. An example is shown in Figure 4 where the addition of *z* = 3 + 2*i* and *w* = 1 + 3*i* is performed graphically on an Argand diagram or, equivalently

*z* + *w* = (3, 2) + (1, 3) = (4, 5)

**Question T1**

If *z* = 4 + 8*i* and *w* = 15 − 12*i*, use an Argand diagram to find the sum, *z* + *w*. Check your answer by means of vector addition using the (*x*, *y*) notation.

**Figure 17** See Answer T1.

Answer T1

The points representing *w* and *z* are plotted on the Argand diagram in Figure 17. Lines drawn from the origin to these points can be considered as vectors which define the sides of a parallelogram and the sum of these vectors is a diagonal of this parallelogram. From the diagram we find that *z* + *w* is 19 − 4*i*. This can be checked numerically

(4, 8) + (15, −12) = (19, −4)

Although complex numbers behave like vectors as far as addition is concerned, when it comes to multiplication and division the two topics diverge. In terms of ordered pairs of real numbers, multiplication of complex numbers can be defined by i

(*a*, *b*) × (*x*, *y*) = [(*ax* − *by*), (*ay* + *bx*)](9)

Although one can introduce complex numbers by this route, which is entirely independent of the symbol *i*, it must be admitted that Equation 9 looks as though it came out of thin air. In practice *i* is a very useful notational convenience which makes Equation 9 look much more natural. The *i* notation is used throughout science and engineering, and even by the purest of pure mathematicians. It is a practice which we follow in *FLAP*.

## 2.2 Polar coordinates

We have seen how it is straightforward to interpret complex addition as vector addition on an Argand diagram. In order to investigate the effect of complex multiplication, try the following question.

**Question T2**

(a) Plot the numbers, 1, 2*i*, −3 −*i* and 2 − *i* on an Argand diagram.

(b) Multiply each of the numbers in part (a) by 2 and plot the resulting points on the same diagram. Suggest a geometric interpretation of multiplication by 2 and check your conjecture by finding the effect of multiplying −1 − *i* by 2.

(c) Repeat parts (a) and (b), but multiply by *i* instead of 2.

(d) Repeat parts (a) and (b), but multiply by 2*i*.

**Figure 18** See Answer T2(a).

**Figure 19** See Answer T2(b).

**Figure 20** See Answer T2(b).

**Figure 21** See Answer T2(c).

Answer T2

(a) The points are plotted on the Argand diagram in Figure 18.

(b) Multiplying each number by 2 gives us

1 × 2 = 2,

2*i* × 2 = 4*i*

(−3 −*i*) × 2 = −6 − 2*i*,

(2 − *i*) × 2 = 4 − 2*i*

These results are plotted in Figure 19.

You can see that all points move along a line through the origin to a position twice as far from the origin.

If we multiply −1 −1*i* by 2 we find

(−1 −1*i*) × 2 = −2 − 2*i*

This point is plotted in Figure 20 and confirms this conclusion.

(c) Multiplying each number by *i* we find:

1 × *i* = *i*,

2*i* × *i* = −2,

(−3 − *i*) × *i* = 1 − 3*i*,

(2 − *i*) × *i* = 1 + 2*i*

Looking at Figure 21 we see that multiplication by *i* produces a rotation of (*π*/2) rad in an anticlockwise direction about the origin. This observation is consistent with

(−1 − *i*) × *i* = 1− *i*

which is also shown on Figure 21.

(d) Multiplying each number by 2*i* we obtain:

1 × 2*i* = 2*i*,

2*i* × 2*i* = −4,

(−3 − *i*) × 2*i* = 2 − 6*i*,

(2 − *i*) × 2*i* = 2 + 4*i*

Looking at Figure 22 we see that multiplication by 2*i* produces a rotation of (*π*/2) rad in an anticlockwise direction about the origin. There is also an increase by a factor of 2 in the distance of the points from the origin.

This observation is consistent with

(−1 − *i*) × 2*i* = 2 − 2*i*

which is also shown on Figure 22.

**Figure 5** Polar coordinates, *r*, *θ*.

The solution to this problem suggests that the geometric interpretation of complex multiplication may involve both a rotation and a change in the distance from the origin, but this is not easy to see if we write complex numbers in the Cartesian form *x* + *iy*.

However, the geometric properties of complex multiplication are quite evident when the complex numbers are expressed in terms of **polar coordinates**. Figure 5 shows a point specified by means of polar coordinates; we can see that the ‘distance’ of the point from the origin is called *r* and *θ* is the angle between the line from the point to the origin and the *x*–axis. Notice that *r* is (by definition) non–negative and that *θ* is conventionally measured anticlockwise. i

Standard results from trigonometry enable us to express the Cartesian coordinates (*x* and *y*) in terms of polar coordinates (*r* and *θ* )

*x* = *r*_{ }cos_{ }*θ*(10)

*y* = *r*_{ }sin_{ }*θ*(11)

This means that we can write a complex number, *z* = *x* + *iy*, in the form

*z* = *r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*)(12)

which is known as the **polar_form_of_a_complex_numberpolar form** (or **polar_representation_of_a_complex_numberpolar representation**) of the complex number, *z*. Examples of complex numbers in polar form are:

2_{ }[cos(*π*/4) + *i*_{ }sin(*π*/4)]

3.5_{ }[cos(*π*/16) + *i*_{ }sin(*π*/16)]

and0.025_{ }[cos(1.05) + *i*_{ }sin(1.05)]

To convert a complex number from polar to Cartesian form we can again use Equations 10 and 11.

For example, *z* = 3_{ }[cos(*π*/4) + *i*_{ }sin(*π*/4)] has *r* = 3 and *θ* = *π*/4. If we substitute these values into Equations 10 and 11, we find

*x* = *r*_{ }cos_{ }*θ* = 3_{ }cos(*π*/4) = $\dfrac{3}{\sqrt{2\os}}$ and *y* = *r*_{ }sin_{ }*θ* = 3_{ }sin(*π*/4) = $\dfrac{3}{\sqrt{2\os}}$

i.e.$z = \dfrac{3}{\sqrt{2\os}} + \dfrac{3}{\sqrt{2\os}}\,i = \dfrac{3}{\sqrt{2\os}}(1+i)$

**Figure 6** The complex number, $z = 1 + \sqrt{3\os}\,i$, in terms of polar coordinates.

It is also straightforward to convert from Cartesian to polar form since the length, *r*, is given by i

and the angle, *θ*, is such that

$\cos\theta = \dfrac{x}{\sqrt{x^2+y^2}}$(14)

$\cos\theta = \dfrac{y}{\sqrt{x^2+y^2}}$(15)

For example, the complex number, $z = 1 + \sqrt{3\os}\,i$ which has *x* = 1 and $y = \sqrt{3\os}$, can be represented in terms of polar coordinates by

$r = \sqrt{x^2+y^2} = \sqrt{1+3\os} = 2$

and since sin(*θ*) = $\dfrac{3}{\sqrt{2\os}}$ and cos(*θ*) = 1/2 we have *θ* = (*π*/3) rad. The position of the number $1 + \sqrt{3\os}\,i$ on an Argand diagram is shown in terms of polar coordinates in Figure 6.

✦ The polar coordinates of three points A, B and C are, respectively,

*r* = 2 *θ* = (*π*/4) rad

*r* = 3 *θ* = (−*π*/3) rad

*r* = 4 *θ* = (5*π*/6) rad

Express the points in Cartesian coordinates.

✧ We can use Equations 10 and 11

*x* = *r*_{ }cos_{ }*θ*(Eqn 10)

*y* = *r*_{ }sin_{ }*θ*(Eqn 11)

to convert from polar coordinates to Cartesian coordinates

For A we have

$x = r\cos\theta = 2\cos(\pi/4) = 2\times 1/\sqrt{2\os} = \sqrt{2\os}$

and$x = r\sin\theta = 2\sin(\pi/4) = 2\times 1/\sqrt{2\os} = \sqrt{2\os}$

For B

$x = r\cos\theta = 3\cos(\pi/3) = 3\times 1/2 = 3/2$

and$x = r\sin\theta = -3\sin(\pi/3) = -3\times \sqrt{3\os}/2 = -3\sqrt{3\os}/2$

For C

$x = r\cos\theta = 4\cos(5\pi/6) = -4\cos(\pi/6) = -4\times \sqrt{3\os}/2 = -2\sqrt{3\os} $

and$x = r\sin\theta = 4\sin(5\pi/6) = 4\sin(\pi/6) = 4\times 1/2 = 2$

One big advantage of the polar representation is that the multiplication of complex numbers is easy when they are expressed in this form. To see this, consider two complex numbers, *z* = *x* + *iy* and *w* = *a* + *ib* for which

*x* = *r*_{ }cos_{ }*θ* and *y* = *r*_{ }sin_{ }*θ*

*a* = *ρ*_{ }cos_{ }*ϕ* and *b* = *ρ*_{ }sin_{ }*ϕ*

Recalling two results from trigonometry:

sin(*α* + *β*) = cos_{ }*α*_{ }sin_{ }*β* + sin_{ }*α*_{ }cos_{ }*β*(Eqn 1) i

cos(*α* + *β*) = cos_{ }*α*_{ }cos_{ }*β* − sin_{ }*α*_{ }sin_{ }*β*(Eqn 2)

we see that the real part of the product *zw* (see Equation 7) is given by

Re(*zw*) = *ax* − *by* = *ρ*_{ }cos_{ }*ϕ* × *r*_{ }cos_{ }*θ* − *ρ*_{ }sin_{ }*ϕ* × *r*_{ }sin_{ }*θ*Re(

*zw*) =

*ρr*

_{ }(cos

_{ }

*θ*

_{ }cos

_{ }

*ϕ*− sin

_{ }

*θ*

_{ }sin

_{ }

*ϕ*)

Re(

*zw*) =

*ρr*

_{ }cos(

*θ*+

*ϕ*)

and the imaginary part is given by

Im(*zw*) = *ay* + *bx* = *ρ*_{ }cos_{ }*ϕ* × *r*_{ }sin_{ }*θ* + *ρ*_{ }sin_{ }*ϕ* × *r*_{ }cos_{ }*θ*

Im(*zw*) = *ρr*_{ }(cos_{ }*ϕ*_{ }sin_{ }*θ* + sin_{ }*ϕ*_{ }cos_{ }*θ*)

Im(*zw*) = *ρr*_{ }sin(*θ* + *ϕ*)

We can summarize these two results by the following rule for multiplying complex numbers in polar form:

*r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*) × *ρ*_{ }(cos_{ }*ϕ* + *i*_{ }sin_{ }*ϕ*) = *ρr*_{ }[cos(*θ* + *ϕ*) + *i*_{ }sin(*θ* + *ϕ*)](16)

So, multiplying a complex number, *w* say, by a complex number with a polar representation *r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*), produces a new complex number which corresponds to the line from the origin to the point representing *w* being first scaled by a factor *r*, then the resulting line being rotated anticlockwise about the origin through an angle *θ*.

**Figure 9** Complex multiplication causing a rotation and a change in distance from the origin.

**Figure 8** Complex multiplication causing a rotation.

**Figure 7** Complex multiplication resulting in a change in distance from the origin.

As an illustration we will start with the complex number *w* = 2_{ }[cos(*π*/4) + *i*_{ }sin(*π*/4)] and first multiply it by 2 in Figure 7, then by [cos(*π*/8) + *i*_{ }sin(*π*/8)] in Figure 8, and finally by 2_{ }[cos(*π*/8) + *i*_{ }sin(*π*/8)] in Figure 9.

Notice that in Figure 9 the line from the origin to the original point is rotated through an angle of (*π*/8) *and* the distance from the origin is doubled.

## 2.3 The modulus of a complex number

Given a complex number, *z* = *x* + *iy*, the **modulus_of_a_complex_numbermodulus** of *z* is defined by

$\lvert\,z\,\rvert = \sqrt{x^2+y^2}$(17)

In polar coordinates, where *x* = *r*_{ }cos_{ }*θ* and *y* = *r*_{ }sin_{ }*θ*, we have

*x*^{2} + *y*^{2} = *r*^{2}_{ }cos^{2}_{ }*θ* + *r*^{2}sin^{2}_{ }*θ* = *r*^{2} (because cos^{2}_{ }*θ* + sin^{2}_{ }*θ* = 1)

So in polar coordinates, the modulus of a complex number is simply the distance from the origin of a point on an Argand diagram. This distance is clearly a non–negative real number.

**Figure 10** See Question T3.

**Question T3**

(a) Figures 10a and 10b show two complex numbers *z*_{1} and *z*_{2}, respectively. Write *z*_{1} and *z*_{2} in the *x* + *iy* form.

(b) Figures 10c and 10d show two complex numbers *z*_{3} and *z*_{4}, respectively. Calculate the modulus of the complex numbers *z*_{3}*z*_{4}.

Answer T3

(a) *z*_{1} = 3_{ }[cos(7*π*/6) + *i*_{ }sin(7*π*/6)] = $3\left(-\dfrac{\sqrt{3\os}}{2}-\dfrac{i}{2}\right) = -\dfrac{3}{2}\left(\sqrt{3\os} + i\right)$

*z*_{2} = 2_{ }[cos(−*π*/3) + *i*_{ }sin(−*π*/3)] = $2\left(\dfrac{1}{2}-\dfrac{i\sqrt{3\os}}{2}\right) = 1 - i\sqrt{3\os}$

(b) |_{ }*z*_{3}*z*_{4}_{ }| = 2 × 3.6056 = 7.2112

(by putting *ρ* = 2 and *r* = 3.6056 into Equation 16),

*r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*) × *ρ*_{ }(cos_{ }*ϕ* + *i*_{ }sin_{ }*ϕ*) = *ρr*_{ }[cos(*θ* + *ϕ*) + *i*_{ }sin(*θ* + *ϕ*)](Eqn 16)

## 2.4 The argument of a complex number

**Figure 6** The complex number,

**Figure 5** Polar coordinates, *r*, *θ*.

If a complex number is written in polar form as *z* = *r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*), then *θ* is known as the **argument** of *z* and is denoted by **argument_of_a_complex numberarg( z)**. For example, if

*z*= 4

_{ }[cos(

*π*/15) +

*i*

_{ }sin(

*π*/15)] then arg(

*z*) is (

*π*/15) rad which can be interpreted geometrically as follows. Consider a typical point representing a complex number,

*z*, on an Argand diagram such as in Figure 5. Then the angle

*θ*made by the line joining the point to the origin with the positive real axis is the argument. Notice that by convention the angle is measured anticlockwise (so that a negative angle would be measured in a clockwise direction). If

*z*= 1 + $\sqrt{3\os}$

*i*then arg(

*z*) is (

*π*/3) rad, as shown in Figure 6.

There is a slight complication in the definition of the argument since a point, *z*, on an Argand diagram does not correspond to a single value of arg(*z*) because we can always add any integer multiple of 2*π* to the value of *θ*; in other words, if we rotate the point about the origin through any number of complete turns we always get back to the same point (see Figure 11). i

**Figure 11** The non–uniqueness of arg(*z*) (shown here as *θ*).

However, if we impose the condition that −*π* < *θ* ≤ *π*i then *θ* is uniquely determined, and with this condition *θ* is known as the **principal value** of the argument of *z*. Notice that whereas the lower limit is greater than −*π*, the upper limit is less than *or equal* to *π*. i

If we have a complex number, *z* = *x* + *iy*, where *x* and *y* are not both zero, then the argument, *θ*, is given by the solution to the following pair of equations

$\cos\theta = \dfrac{x}{\sqrt{x^2+y^2}}$(18)

$\sin\theta = \dfrac{y}{\sqrt{x^2+y^2}}$(19)

From Equations 18 and 19 we obtain

tan_{ }*θ* = *y*/*x*(20)

and while it is possible to use Equation 20 to find the angle *θ*, this must be done with some care as the following example shows. (Alternatively we can use the Equations 18 and 19, see Solution B to Example 1).

### Example 1

Find the principal value of the argument of the complex number *z* = −1 + *i*.

#### Solution A

**Figure 12** The complex number, *z* = −1 + *i*, plotted on an Argand diagram.

In this case *x* = −1 and *y* = 1, and from Equation 20 we have tan_{ }*θ* = −1. In order to find *θ* your first thought might be to set your calculator to radian mode and to evaluate arctan(−1), which will give you the approximate value −0.785 398 rad for *θ*.

However, this is *not* a value of the argument, and certainly not the principal value, as you can easily see if you plot the point −1 + *i* on an Argand diagram, see Figure 12. The correct answer is approximately 2.356 194 rad (or more precisely (3*π*/4) rad).

The essential point to realize here is that Equation 20 does not determine the angle *θ* uniquely because there are generally two angles in the range −*π* < *θ* ≤ *π* that correspond to a given value of the tangent, and these angles differ by *π* radians. The difficulty is quite easy to resolve if we always draw a diagram (such as Figure 12) when calculating a value for arg(*z*). In this case we would obtain the value −00.785 398 rad from the calculator as before, then we see from the diagram that we must add *π* radians to obtain the principal value

arg(*z*) ≈ −0.785 0398 + 3.141 1593 = 2.356 0194 rad

For this particular value of *z* you may be able to avoid the use of the calculator if you can see that arg(*z*) = (3*π*/4) rad directly from the figure.

#### Solution B

Alternatively we can use Equations 18 and 19 to obtain $\cos\theta = -1/\sqrt{2\os} \quad\text{and}\quad\sin\theta = -1/\sqrt{2\os}$. The only angles in the range −*π* < *θ* ≤ *π* that satisfy the first equation are *θ* = (3*π*/4) rad or *θ* = (−3*π*/4) rad, while the only angles that satisfy the second equation are *θ* = (*π*/4) rad or *θ* = (3*π*/4) rad. Thus the required angle is *θ* = (3*π*/4) rad.

Notice that the argument of *z* = *x* + *iy* is not defined when *x* and *y* are *both* zero; in other words, the argument of *z* = 0 is not defined.

The polar representation of a complex number is not unique. For example, both $\sqrt{2\os}[\cos(\pi/4) + i\,\sin(\pi/4)]\;\text{and}\;\sqrt{2\os}[\cos(9\pi/4) + i\,\sin(9\pi/4)]$ represent the same complex number 1 + *i*. We can, however, make the representation unique if we insist that the argument takes its principal value. Notice that $-2[\cos(\pi/)3 + i\,\sin(\pi/3)]$ is *not* in polar form. In fact from Equation 16,

*r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*) × *ρ*_{ }(cos_{ }*ϕ* + *i*_{ }sin_{ }*ϕ*) = *ρr*_{ }[cos(*θ* + *ϕ*) + *i*_{ }sin(*θ* + *ϕ*)](Eqn 16)

−2_{ }[cos(*π*/3) + *i*_{ }sin(*π*/3)] = 2_{ }(cos_{ }*π* + *i*_{ }sin_{ }*π*)[cos(*π*/3) + *i*_{ }sin(*π*/3)] = 2_{ }[cos(4*π*/3) + *i*_{ }sin(4*π*/3)]

and this final result *is* in polar form.

**Figure 1** See Question R3.

**Question T4**

For each of the following complex numbers, find the principal value of the argument:

(a) $-1 + \sqrt{3\os}\,i$, (b) 1 - $\sqrt{3\os}\,i$, (c) $\sqrt{3\os} + i$, (d) 3 + 2*i*.

[Hint: You may find Figure 1 helpful.]

**Figure 23** See Answer T4(a).

**Figure 24** See Answer T4(b).

**Figure 25** See Answer T4(c).

Answer T4

You will find it is useful to draw an Argand diagram for such problems.

(a) The position of the point representing $-1 + \sqrt{3\os}\,i$ on an Argand diagram is shown in Figure 23. The angle made by the line joining the point to the origin with the negative real axis is (*π*/3) rad since cos(*π*/3) = 1/2. The principal value of arg(*z*) is the angle made with the *positive* real axis which we can see from the diagram to be (2*π*/3) rad.

(b) The position of the point representing $1 - \sqrt{3\os}\,i$ on an Argand diagram is shown in Figure 24. The angle made by the line joining the point to the origin with th positive real axis is (*π*/3) rad since cos(*π*/3) = 1/2. The principal value of arg(*z*) is −*π*/3 since the angle is measured anticlockwise.

(c) The position of the point representing $\sqrt{3\os} + i$ on an Argand diagram is shown in Figure 25. The angle made by the line joining the point to the origin with the positive real axis is (*π*/6) rad since cos(*π*/6) = $\sqrt{3\os}$/2. This is the principal value of arg(*z*).

(d) A sketch shows that the complex number 3 + 2*i* lies in the first

quadrant and, from Equation 20,

tan_{ }*θ* = *y*/*x*(Eqn 20)

tan_{ }*θ* = 2/3. It therefore follows that *θ* ≈ 0.5880 rad.

## 2.5 The exponential form

You should be familiar with the following functions, and their series expansions i

$\displaystyle {\rm e}^x = \sum_{n=0}^{\infty} \dfrac{x^n}{n!} = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dots$(Eqn 3) i

$\displaystyle \sin x = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \dots$(Eqn 4)

$\displaystyle \cos x = \sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n}}{(2n)!} = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots$(Eqn 5)

Each of these functions may be extended so that they apply to a complex variable *z* (rather than the real variable *x*), but, since real numbers are just a special case of complex numbers, the new functions must agree with the old ones in the special case when *z* is real. For example, in order to define e^{z} we simply replace *x* by *z* in Equation 3,

$\displaystyle {\rm e}^z = \sum_{n=0}^{\infty} \dfrac{z^n}{n!} = 1 + \dfrac{z}{1!} + \dfrac{z^2}{2!} + \dfrac{z^3}{3!} + \dots$(21) i

Now let us consider the special case of *z* = *iθ*, where *θ* is real and therefore *z* is imaginary

$\displaystyle {\rm e}^{i\theta} = 1 + \dfrac{{i\theta}}{1!} + \dfrac{({i\theta})^2}{2!} + \dfrac{({i\theta})^3}{3!} + \dfrac{({i\theta})^4}{4!} + \dfrac{({i\theta})^5}{5!} + \dots$

This expression can be simplified by using *i*^{2} = −1 to give

$\displaystyle {\rm e}^{i\theta} = 1 + \dfrac{{i\theta}}{1!} - \dfrac{\theta^2}{2!} + \dfrac{i\theta^3}{3!} - \dfrac{\theta^4}{4!} + \dfrac{i\theta^5}{5!} - \dots$

We can see that the terms are alternately real and imaginary, so it is useful to split the series into two

$\displaystyle {\rm e}^{i\theta} = \left(1 - \dfrac{\theta^2}{2!} + \dfrac{\theta^4}{4!} - \dfrac{\theta^6}{6!} + \dots\right) + i\left(\dfrac{{\theta}}{1!} - \dfrac{\theta^3}{3!} + \dfrac{\theta^5}{5!} - \dfrac{\theta^7}{7!} + \dots\right)$

If we compare the right–hand side of this equation with the series for sin_{ }*θ* and cos_{ }*θ* we see that e^{iθ} can be written as

e^{iθ} = cos_{ }*θ* + *i*_{ }sin_{ }*θ* (*Euler’s formula*)(22)

This important formula, known as **Euler’s formula**, gives us the real and imaginary parts of e^{iθ}. i

**Question T5**

Plot e^{inπ/8} on an Argand diagram for *n* = 0, 1, 2, ... 15. What geometrical shape do you think would describe the position of the points representing *z* = e^{iθ} where *θ* can take any real value?

**Figure 26** See Answer T5.

Answer T5

There are two ways to answer this question.

(a) Euler’s formula enables us to write

*z* = e^{nπi/8} = cos(*nπ*/8) + *i*_{ }sin(*nπ*/8)

Then use a calculator to evaluate the sine and cosine for *n* = 0, 1, 2, ... 15 and plot the points on an Argand diagram, as in Figure 26. Notice that the points all lie on a circle of radius, 1, with the origin as the centre.

The following alternative method is somewhat easier.

(b) Notice that if we compare

*z* = e^{nπi/8} = cos(*nπ*/8) + *i*_{ }sin(*nπ*/8)

with the polar form

*z* = *r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*)

then we have *r* = 1 and arg(*z*) = *nπ*/8. The value of *r* tells us that all points are a distance of 1 from the origin; that is all points lie on a circle of radius 1 centred on the origin. The *θ* coordinate is the value of arg(*z*) for each *n* and points can therefore be plotted directly on the Argand diagram.

The second approach also shows that all points representing *z* = e^{iθ}, where *θ* takes any real value, lie on a circle of radius 1, whose centre is the origin. (This circle is sometimes known as the *unit circle*.)

If we put *θ* = *π* in Euler’s formula then, since cos(*π*) = −1 and sin(*π*) = 0, we find i

This identity is quite remarkable since it relates:

- the numerical constant e, which originates from problems of growth and decay;
- the numerical constant
*π*, which originates from the ratio of the circumference to the diameter of a circle; - the number 1, which has the special property that 1 ×
*n*=*n*for any number*n*; - the symbol
*i*, which has the property*i*^{2}= −1 and was originally introduced in order to solve equations such as*x*^{2}+ 1 = 0.

**Question T6**

Use Euler’s formula to prove the following important relations (Equations 24 and 25):

$\sin\theta = \dfrac{{\rm e}^{i\theta} - {\rm e}^{-i\theta}}{2i}$(24)

$\cos\theta = \dfrac{{\rm e}^{i\theta} + {\rm e}^{-i\theta}}{2}$(25)

Answer T6

We can use Euler’s formula and the fact that cos(−*θ*) = cos_{ }*θ* and sin(−*θ*) = −sin_{ }*θ* to obtain

e^{iθ} = cos_{ }*θ* + *i*_{ }sin_{ }*θ*

e^{−iθ} = cos_{ }*θ* − *i*_{ }sin_{ }*θ*

Addition of the two identities gives e^{iθ} + e^{−iθ} = 2_{ }cos_{ }*θ*

and therefore

$\cos\theta = \dfrac{{\rm e}^{i\theta}+{\rm e}^{-i\theta}}{2}$

Subtraction of the two identities gives e^{iθ} − e^{−iθ} = 2*i*_{ }sin_{ }*θ*

$\sin\theta = \dfrac{{\rm e}^{i\theta}-{\rm e}^{-i\theta}}{2i}$

Now let us consider a general complex number, *z* = *x* + *iy*, and write it in terms of polar coordinates by substituting *x* = *r*_{ }cos_{ }*θ* and *y* = *r*_{ }sin_{ }*θ*

*z* = *x* + *iy* = *r*_{ }cos_{ }*θ* + *ir*_{ }sin_{ }*θ* = *r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*)

which can be written more compactly, using Euler’s equation, to obtain

*z* = *r*_{ }e^{iθ}(26)

which is known as the **exponential_form_of_a_complex_numberexponential form** (or **exponential_representation_of_a_complex_numberexponential representation**) of a complex number. Notice that *r* is (by definition) non-negative.

**Question T7**

Find the exponential representation for the following complex numbers:

(a) $\sqrt{2\os}(1+ i)$, (b) $3(1+ i\sqrt{3\os})$, (c) $2(\sqrt{3\os} + i)$, (d) 0.2 + 2.3*i*.

**Figure 1** See Answer T7.

Answer T7

Initially we want to write each number in the form

*z* = *r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*)

and we can do this by ensuring that the sum of the squares of the real and imaginary parts within the parentheses is 1. This can be achieved in each case by dividing the complex number in parentheses by its modulus; for example, the complex number (1 + *i*) has modulus $\sqrt{2\os}$ and we therefore write (1 + *i*) = $\sqrt{2\os}\left(\dfrac{1}{\sqrt{2\os}}+\dfrac{i}{\sqrt{2\os}}\right)$

(a) $\sqrt{2\os}(1+i) = 2\left(\dfrac{1}{\sqrt{2\os}}+\dfrac{i}{\sqrt{2\os}}\right)$

The value of *r* is 2 and *θ* is *π*/4 rad since cos(*π*/4) = 1/$\sqrt{2\os}$ and sin(*π*/4) = 1/$\sqrt{2\os}$. To check these values either use a calculator or, much better, consider a right–angled triangle with two sides of equal length, as in Question R4. So the exponential form is 2_{ }e^{iπ/4}.

(b) $3(1+i\sqrt{3\os}) = 6\left(\dfrac{1}{2}+\dfrac{i\sqrt{3\os}}{2}\right)$

The value of *r* is 6. To obtain *θ* we need to solve the equations cos_{ }*θ* = 1/2 and sin_{ }*θ* = $\sqrt{3\os}$/2, and from Figure 1 (see Question R3) we see that the result is *θ* = (*π*/3) rad. So the exponential form is 6_{ }e^{iπ/3}.

(c) $2(\sqrt{3\os}+i) = 4\left(\dfrac{\sqrt{3\os}}{2}+\dfrac{i}{2}+\right)$

We have *r* = 4 and *θ* is the solution of cos_{ }*θ* = $\sqrt{3\os}$/2 and sin_{ }*θ* = 1/2. From Figure 1 (see Question R3) we get *θ* = (*π*/6) rad and therefore the exponential form is 4_{ }e^{iπ/6}.

(d) The modulus of 0.2 + 2.3*i* is $\sqrt{(0.2)^2+(2.3)^2} \approx 2.309$ and since the point lies in the first quadrant its argument is arctan(2.3/0.2) ≈ 1.484 rad. Hence 0.2 + 2.3*i* ≈ 2.309_{ }e^{1.484i}.

✦ Express the complex numbers *i* and 2 in exponential form.

✧ From Euler’s formula *i* = e^{iπ/2} (because sin(*π*/2) = 1 and cos(*π*/2) = 0) and 2 = 2 × 1 = 2_{ }e^{i0} (because sin(0) = 0 and cos(0) = 1).

## 3 Arithmetic of complex numbers

## 3.1 Products and quotients

Multiplication of complex numbers in exponential form is very easy, however in this form addition is more difficult.

### Products

One important property of real powers of any real quantity is the identity

*a*^{ s}*a*^{ t} = *a*^{ s + t}(27) i

which we will assume is also true if any (or all) of the quantities *a*, *s* and *t* are complex. This identity allows us to multiply complex numbers easily in exponential form since, if *z* = *r*_{ }e^{iθ} and *w* = *ρ*_{ }e^{iϕ}, then the product is given by

*r*_{ }e^{iθ} × *ρ*_{ }e^{iϕ} = *rρ*_{ }e^{i (θ + ϕ)}(28)

which is closely related to the result quoted in Equation 16,

*r*_{ }(cos_{ }*θ* + *i*_{ }sin_{ }*θ*) × *ρ*_{ }(cos_{ }*ϕ* + *i*_{ }sin_{ }*ϕ*) = *ρr*_{ }[cos(*θ* + *ϕ*) + *i*_{ }sin(*θ* + *ϕ*)](Eqn 16)

Notice that in Equation 28 the moduli i of the two complex numbers are multiplied whereas the arguments are added (as in Equation 16).

**Question T8**

Find *zw* in exponential form (choosing the principal value of the argument in each case) for each of the following:

(a) *z* = 2_{ }e^{iπ/4} *w* = 2_{ }e^{iπ/4}

(b) *z* = 3_{ }e^{iπ} *w* = 2_{ }e^{iπ/4}

(c) *z* = 2_{ }e^{iπ/16} *w* = ½ e^{−iπ/16}

Answer T8

(a) *zw* = 2_{ }e^{iπ/4} × 2_{ }e^{iπ/4} = 4_{ }e^{iπ/2}

(b) *zw* = 3_{ }e^{iπ} × 2_{ }e^{iπ/4} = 6_{ }e^{5iπ/4} = 6_{ }e^{5iπ/4} × e^{−2iπ} = 6_{ }e^{−3iπ/4}

Notice that we have used the result e^{−2iπ} = 1 to make the argument of *zw* consistent with the range −*π* < arg(*zw*) ≤ *π*.

(c) *zw* = 2_{ }e^{iπ/16} × ½ e^{−iπ/16} = e^{0} = 1

### Quotients

For real numbers we have the general result that $\dfrac{a^m}{b^n} = a^mb^{-n}$ where *b* is non-zero.

Extending this result to complex numbers, we find that simplifying a quotient of two complex numbers in exponential form is a straightforward variation of multiplication. So if *z* = *r*_{ }e^{iθ} and *w* = *ρ*_{ }e^{iϕ}, then *z*/*w* is given by $\dfrac{z}{w} = \dfrac{r{\rm e}^{i\theta}}{\rho{\rm e}^{i\phi}} = \dfrac{r}{\rho}{\rm e}^{i(\theta-\phi)}$ that is, we divide the moduli and subtract the arguments.

**Question T9**

Find *z*/*w* in exponential form (choosing the principal value of the argument in each case) for each of the following:

(a) *z* = 2_{ }e^{iπ/4} *w* = 4_{ }e^{−iπ/8}

(b) *z* = 3_{ }e^{iπ}/4 *w* = 2_{ }e^{iπ/2}

(c) *z* = 2_{ }e^{−iπ/16} *w* = 2_{ }e^{iπ/16}

Answer T9

(a) $\dfrac{z}{w} = \dfrac{2{\rm e}^{i\pi/4}}{4{\rm e}^{-i\pi/8}} = \dfrac{1}{2}{\rm e}^{3i\pi/8}$

(b) $\dfrac{z}{w} = \dfrac{3{\rm e}^{i\pi/4}}{2{\rm e}^{i\pi/2}} = \dfrac{3}{2}{\rm e}^{-i\pi/4}$

(c) $\dfrac{z}{w} = \dfrac{2{\rm e}^{-i\pi/16}}{2{\rm e}^{i\pi/16}} = {\rm e}^{-i\pi/8}$

## 3.2 Sums and differences

Suppose that we are given two complex numbers in exponential form, *a*_{ }e^{iα} and *b*_{ }e^{iβ}, we can certainly write their sum in exponential form

*a*_{ }e^{iα} + *b*_{ }e^{iβ} = *c*_{ }e^{iγ}(29) i

but it is not particularly easy to find the values of *c* and *γ* directly. First we will show you a direct method of finding the sum, then an alternative method which is often easier.

### Method 1

To find *c* we can write

*c*^{2} = (*c*_{ }e^{iγ})(*c*_{ }e^{−iγ}) = (*a*_{ }e^{iα} + *b*_{ }e^{iβ})(*a*_{ }e^{−iα} + *b*_{ }e^{−iβ})*c*^{2} = *a*^{2} + *b*^{2} + *ab*_{ }[e^{i (α − β)} + e^{−i (α − β)}]*c*^{2} = *a*^{2} + *b*^{2} + 2*ab*_{ }cos(*α* − *β*)

So we have an equation for *c*

$c= \sqrt{\smash[b]{a^2+b^2+2ab\cos(\alpha-\beta)}}$(30) i

To find *γ* we can use Euler’s formula to write e^{iα}, e^{iβ} and e^{iγ} in polar form, and Equation 29 then becomes

*a*_{ }(cos_{ }*α* + *i*_{ }sin_{ }*α*) + *b*_{ }(cos_{ }*β* + *i*_{ }sin_{ }*β*) = *c*_{ }(cos_{ }*γ* + *i*_{ }sin_{ }*γ*)

and equating real and imaginary parts gives us:

$\cos\gamma = \dfrac{a\cos\alpha + b\cos\beta}{c}$

$\sin\gamma = \dfrac{a\sin\alpha + b\sin\beta}{c}$

So if we are given *a*, *b*, *α* and *β* we can first find *c*, then cos_{ }*γ* and sin_{ }*γ*. Then the condition, −*π* < *γ* ≤ *π* uniquely fixes *γ*.

✦ Given that *z* = *a*_{ }e^{iα} = e^{2iπ/3} and *w* = *b*_{ }e^{iβ} = e^{iπ/3}, find *z* + *w* in exponential form.

✧ If *z* + *w* = *c*_{ }e^{iγ}

then *c*^{2} = *a*^{2} + *b*^{2} + 2*ab*_{ }cos(*α* −*β*)

then *c*^{2} = 1^{2} + 1^{2} + 2 × 1 × 1 × cos(2*π*/3 − *π*/3)

then *c*^{2} = 1^{2} + 1^{2} + 2 × 1 ×1 × cos(*π*/3) = 3

which gives $c = \sqrt{3\os}$.

The angle, *γ*, can be obtained from

$\cos\gamma = \dfrac{a\cos\alpha+b\cos\beta}{c}$

$\phantom{\cos\gamma }= \dfrac{\cos(2\pi/3)+\cos(\pi/3)}{\sqrt{3\os}}$

$\phantom{\cos\gamma }= \dfrac{1}{\sqrt{3\os}}\left(-\dfrac{1}{2}+\dfrac{1}{2}\right) = 0$

and$\sin\gamma = \dfrac{a\sin\alpha+b\sin\beta}{c}$

$\phantom{\cos\gamma }= \dfrac{\sin(2\pi/3)+\sin(\pi/3)}{\sqrt{3\os}}$

$\phantom{\cos\gamma }= \dfrac{1}{\sqrt{3\os}}\left(\dfrac{\sqrt{3\os}}{2}+\dfrac{\sqrt{3\os}}{2}\right) = 1$

The only angle *γ* in the range −*π* < *γ* ≤ *π* for which cos_{ }*γ* = 0 and sin_{ }*γ* = 1 is *γ* = *π*/2 and therefore

e^{2iπ/3} + e^{iπ/3} = 3_{ }e^{iπ/24}

### Method 2

The alternative method of finding the sum in exponential form relies on the fact that addition in Cartesian form is very straightforward. So if *z* and *w* are given in exponential form, we first convert them into Cartesian form, then find the sum, and finally convert the answer back into exponential form. We can use the previous example to illustrate the method.

Given that *z* = e^{2iπ/3} and *w* = e^{iπ/3}, we can use Euler’s formula (Equation 22)

e^{iθ} = cos_{ }*θ* + *i*_{ }sin_{ }*θ*(Eqn 22)

to write *z* and *w* in Cartesian form

$z = \dfrac{-1+i}{\sqrt{3\os}}{2}$ and $w = \dfrac{1+i}{\sqrt{3\os}}{2}$

from which it is straightforward to find the sum

$z + w = \dfrac{-1+i}{\sqrt{3\os}}{2} +\dfrac{1+i}{\sqrt{3\os}}{2} = i\sqrt{3\os}$

Since $i\sqrt{3\os} = \sqrt{3\os}\,{\rm e}^{i\pi/2}$ this result for *z* + *w* agrees with our previous result, but you should now be convinced that addition (and subtraction) are much easier using the Cartesian rather than exponential form of complex numbers.

**Question T10**

Express e^{iπ} − e^{iπ/2} in exponential form.

Answer T10

${\rm e}^{i\pi} - {\rm e}^{i\pi/2} = -1 - i = \sqrt{2\os}{\rm e}^{-3i\pi/4}$

**Question T11**

Use the exponential representation to show that for any complex numbers *z*_{1} and *z*_{2}

|_{ }*z*_{1}_{ }| + |_{ }*z*_{2}_{ }| ≥ |_{ }*z*_{1} + *z*_{2}_{ }| i

[Hint: Use Equation 30 and the fact that the cosine of any angle is less than or equal to one.]

$c= \sqrt{\smash[b]{a^2+b^2+2ab\cos(\alpha-\beta)}}$(Eqn 30)

Answer T11

We write *z*_{1} and *z*_{2} in exponential form as *z*_{1} = *a*_{ }e^{iα} and *z*_{2} = *b*_{ }e^{iβ}, so we immediately have |_{ }*z*_{1}_{ }| + |_{ }*z*_{2}_{ }| = *a* + *b*. Now let *z*_{1} + *z*_{2} = *c*_{ }e^{iγ}, then, from Equation 30, we have

$c = \sqrt{\smash[b]{a^2+b^2+2ab\cos(\alpha-\beta)}}$(Eqn 30)

But cos(*α* − *β*) ≤ 1, so we have

|_{ }*z*_{1} + *z*_{2}_{ }| = *c* ≤ $\sqrt{a^2+b^2+2ab}$ = *a* + *b* = |_{ }*z*_{1}_{ }| + |_{ }*z*_{2}_{ }|

and therefore |_{ }*z*_{1}_{ }| + |_{ }*z*_{2}_{ }| ≥ |_{ }*z*_{1}+ *z*_{2}_{ }|.

## 3.3 Powers, roots and reciprocals

The meaning of an expression such as (1 + 2*i*)^{3} is clear enough, i however the meaning of an expression such as (1 + 2*i*)^{ π} is by no means obvious. In this subsection we will attempt to attach a sensible meaning to such expressions. The following results are certainly true for real numbers *a* and *b*

(*uv*)^{ a} = *u*^{ a}*v*^{ a} and (*u*^{ a})^{ b} = *u*^{ ab}

and we assume that they also hold for complex numbers. So to raise the complex number *z* = *r*_{ }e^{iθ} to the power *α* we have

*z* = (*r*_{ }e^{iθ})^{ α} = *r*^{ α}_{ }e^{iαθ}(31)

For example, if *z* = 2_{ }e^{iπ/4} then

*z*^{2} = 4_{ }e^{iπ/2} = 4_{ }[cos(*π*/2) + *i*_{ }sin(*π*/2)] = 4*i*

We can check this result by first putting *z* into Cartesian form

*z* = 2_{ }e^{iπ/4} = 2_{ }[cos(*π*/4) + *i*_{ }sin(*π*/4)] = $\dfrac{2}{\sqrt{2\os}}(1 + i) = \sqrt{2\os}(1+i)$

and then evaluating *z*^{2} to give *z*^{2} = 2_{ }(1 + 2*i* + *i*^{2}) = 4*i*.

**Question T12**

Find *z*^{2} and *z*^{3}, where *z* = 2_{ }e^{iπ/3}. Plot *z* and *z*^{2} and *z*^{3} on an Argand diagram and use it to explain the movement of the point representing *z*^{ n} for successive integer values of *n*.

**Figure 27** See Answer T12.

Answer T12

*z*^{2} = (2_{ }e^{iπ/3})^{2} = 2^{2}e^{2iπ/3} = 4_{ }e^{2iπ/3}

*z*^{3} = (2_{ }e^{iπ/3})^{3} = 2^{3}e^{3iπ/3} = −8_{ }e^{3iπ/3} = −8 (since e^{iπ} = −1)

If we plot the result of multiplying any complex number by *z* on an Argand diagram, then the distance from the origin will increase by a factor of 2 and the point will rotate about the origin through an angle of (*π*/3) rad. Therefore multiplying by *z*^{2} will increase the distance by a factor of 4 and rotate the point through an angle of (2*π*/3) rad. This is confirmed by Figure 27.

It is certainly possible to calculate powers of a complex number in Cartesian form, but it must be done with some care if we are to keep the algebra under control.

✦ Express (1 + *i*)^{8} in Cartesian form.

✧ (1 + *i*)^{2} = (1 + 2*i* + *i*^{2}) = 2*i*

so that(1 + *i*)^{4} = (2*i*)^{2} = −4

and(1 + *i*)^{8} = (−4)^{2} = 16.

Very often a better method would be to express the complex number in exponential form, then raise it to the power, and finally convert the answer back into Cartesian form. Using this approach to evaluate (1 + *i*)^{8} we have

$(1 + i) = \sqrt{2\os}\,{\rm e}^{i\pi/4}$ so that $(1 + i)^8 = (\sqrt{2\os}\,{\rm e}^{i\pi/4})^8 = (\sqrt{2\os})^8\,{\rm e}^{2\pi i} = 16$

**Question T13**

Evaluate i $(1 + \sqrt{3\os}\,i)^3$ and hence express $(1 + \sqrt{3\os}\,i)^{10}$ in Cartesian form. Also express $z = 1 + \sqrt{3\os}\,i$ in exponential form and use this result to evaluate $(1 + \sqrt{3\os}\,i)^{10}$.

Answer T13

First we have $(1 + \sqrt{3\os}\,i)^2 = (1 - 3 + 2\sqrt{3\os}\,i) = -2 + \sqrt{3\os}\,i$

We then use the result to calculate $(1 + \sqrt{3\os}\,i)^3$

$(1 + \sqrt{3\os}\,i)^3 = (-2 + 2\sqrt{3\os}\,i)(1+ 3i) = -2\times 1- 2\times 3+i\,(-2\times 3 + 2\sqrt{3\os}\,i\times 1) = -8$

So we have

$(1 + \sqrt{3\os}\,i)^{10} = (1 + \sqrt{3\os}\,i)^9(1 + \sqrt{3\os}\,i) = (-8)^3(1 + \sqrt{3\os}\,i) = -512(1 + \sqrt{3\os}\,i)$

Now compare this calculation with the following which uses the exponential form of a complex number.

Using cos(*π*/3) = 1/2 and sin(*π*/3) = $\sqrt{3\os}$/2 we have

$z = 1+\sqrt{3\os} = {\rm e}^{i\pi/3}$

and therefore

$z^{10} = 2^{10}{\rm e}^{10i\pi/3} = 2^{10}{\rm e}^{-2i\pi/3} = 2^{10}[\cos(2\pi/3) -i\sin(2\pi/3)]$

$\phantom{z^{10} }= 2^{10}(-1/2-i\sqrt{3\os}/2) = -512(1 + \sqrt{3\os}\,i)$

This is consistent with the previous method but is a little easier.

**Question T14**

The complex number, *u*, is defined by

*u* = (1.1)_{ }e^{2πi/15}

Use an Argand diagram to plot *u*^{ n} for *n* = 1, 2, 3, ... 15. What sort of curve would you expect to get if larger (integer) values of *n* were plotted and the points joined by a smooth curve?

**Figure 28** See Answer T14.

Answer T14

The argument of *u*^{ n} is 2*πn*/15. The point representing *u*^{ n} for *n* = 1 lies on a line from the origin which makes an angle of (2*π*/15) rad with the positive real axis. For any integer *n* this angle increases to (2*nπ*/15) rad.

For *n* = 1, the distance of the point from the origin is 1.1, and in general the distance is (1.1)^{ n}. The result is the series of points shown in Figure 28. Notice that the points lie on a spiral curve.

The following example illustrates an area of physics that uses complex numbers.

✦ The propagation constant of a cable carrying an *alternating current* of *angular frequency**ω* is given by

$\sigma = \sqrt{(R+i\omega L)(G+i\omega C)\os}$ i

where *R*, *L*, *G* and *C* are, respectively, the *resistance*, *inductance*, *conductance* and *capacitance* of the cable i

Evaluate *σ* if *R* = 90 Ω, *L* = 0.002 H, *G* = 5 × 10^{−5} S, *C* = 0.05 × 10^{−6} F and *ω* = 5000 s^{−1}.

✧ *R* + *iω*_{ }*L* = 90 Ω + [(5000) × (0.002)*i*] Ω = (90 + 10*i*) Ω

*G* + *iωC* = [5 × 10^{−5} + (5000) × (0.05 × 10^{−6})*i*] S = (5 + 25*i*) × 10^{−5} S

It follows that

(*R* + *iωL*)(*G* + *iωC*) = (90 + 10*i*) Ω × (5 + 25*i*) × 10^{−5} Ω^{−1}

(*R* + *iωL*)(*G* + *iωC*) = (0.2 + 2.3*i*) × 10^{−2} ≈ (2.309) e^{1.484i} × 10^{−2} (see Question T7(d))

Therefore$\sigma = \sqrt{(R+i\omega L)(G+i\omega C)}$ ≈ [(2.309)_{ }e^{1.484} *i*]^{1/2} × 10^{−1} ≈ (0.1519)_{ }e^{0.742 i}

### Complex powers

The meaning of a complex power of a complex number becomes clear if we use the exponential, rather than Cartesian, form. For example

*z* = (e^{iπ/4})^{ i} = e

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