Complex numbers: Represent complex numbers in a form appropriate to the context

# Unit 3: Represent complex numbers in polar form

Dylan Busa

### Unit outcomes

By the end of this unit you will be able to:

• Express complex numbers in polar form.
• Convert complex numbers from standard/rectangular form to polar form.
• Convert complex numbers from polar form to standard/rectangular form.

## What you should know

Before you start this unit, make sure you can:

• Represent complex numbers using an Argand diagram. Refer to unit 2 in this subject outcome if you need help with this.
• Find the modulus of a complex number. Refer to unit 2 in this subject outcome if you need help with this.
• Find the argument of a complex number. Refer to unit 2 in this subject outcome if you need help with this.

## Introduction

We saw in the previous unit that we can represent complex numbers on the complex plane. These are referred to as Argand diagrams. Figure 1 shows the complex number $\scriptsize z=2-3i$ represented in this way, where the complex number is represented as the point $\scriptsize (x,y)=(2,-3)$.

We also saw that we can calculate the length of the line joining the point representing a complex number with the origin. This is called the absolute value or modulus of the complex number and is represented as $\scriptsize \left| z \right|$. The modulus of $\scriptsize z=2-3i$ is calculated as follows:

\scriptsize \begin{align*}\left| z \right| & =\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\ & =\sqrt{{{{2}^{2}}+{{{(-3)}}^{2}}}}\\ & =\sqrt{{4+9}}\\ & =\sqrt{{13}}\end{align*}

We noted that it is not enough to know just the modulus of a complex number. We also need to know the angle that the line representing the modulus makes with the positive x-axis (or positive real axis). This angle is called the argument.

We can calculate the argument using either of the following trigonometric ratios:
$\scriptsize \sin \theta =\displaystyle \frac{y}{r}$ or $\scriptsize \cos \theta =\displaystyle \frac{x}{r}$ where $\scriptsize r$ (the radius of the circle with centre the origin, passing through the point $\scriptsize (x,y)$ representing the complex number) has the same value as $\scriptsize \left| z \right|$.

In the case above the argument is calculated as:

\scriptsize \begin{align*}\sin \theta & =\displaystyle \frac{y}{r}=\displaystyle \frac{{-3}}{{\sqrt{{13}}}}\\ & \therefore \theta ={{303.69}^\circ}\quad \text{The point is in the fourth quadrant}\end{align*}

With all of this knowledge, we are ready to work with complex numbers in polar form.

## Complex numbers in polar form

We say that $\scriptsize z=2-3i$ is written in standard or rectangular form where the number is expressed in terms of a real and an imaginary component. The polar form of a complex number expresses the number in terms of its modulus and argument.

Suppose we have a complex number $\scriptsize z=x+yi$. We can represent this with an Argand diagram as shown in Figure 2.

Now we know that $\scriptsize \cos \theta =\displaystyle \frac{x}{r}$ and $\scriptsize \sin \theta =\displaystyle \frac{y}{r}$. Therefore, $\scriptsize x=r\cos \theta$ and $\scriptsize y=r\sin \theta$.

So, the point $\scriptsize (x,y)$ has coordinates given by $\scriptsize x=r\cos \theta$ and $\scriptsize y=r\sin \theta$ where $\scriptsize r=\sqrt{{{{x}^{2}}+{{y}^{2}}}}$. Therefore:

\scriptsize \begin{align*}z & =x+yi\\ & =r\cos \theta +(r\sin \theta )i\\ & =r(\cos \theta +i\sin \theta )\end{align*}

### Take note!

We often use the abbreviation $\scriptsize r\text{cis}\theta$ to represent $\scriptsize r(\cos \theta +i\sin \theta )$.

The polar form of a complex number:

\scriptsize \displaystyle \begin{align*}z & =x+yi\\ & =r\cos \theta +(r\sin \theta )i\\ & =r(\cos \theta +i\sin \theta )\\ & =r\text{cis}\theta \end{align*}

where $\scriptsize r=\left| z \right|$ and $\scriptsize \theta$ is the argument.

### Example 3.1

Find the polar form of $\scriptsize z=4+4i$.

Solution

Step 1: Find the value of $\scriptsize r$ (or $\scriptsize \left| z \right|$ or the modulus)

\scriptsize \begin{align*}r & =\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\ & =\sqrt{{{{4}^{2}}+{{4}^{2}}}}\\ & =\sqrt{{16+16}}\\ & =\sqrt{{32}}\quad 32=16\times 2\\ & =4\sqrt{2}\end{align*}

Step 2: Find the value of $\scriptsize \theta$ (the argument)

\scriptsize \begin{align*}\cos \theta & =\displaystyle \frac{x}{r}=\displaystyle \frac{4}{{4\sqrt{2}}}=\displaystyle \frac{1}{{\sqrt{2}}}\\ & \therefore \theta ={{45}^\circ}\end{align*}

In this case, we used the special angle of $\scriptsize {{45}^\circ}$. We know the point is in the first quadrant so we do not need to transform the angle to another quadrant.

Note that we could also have used $\scriptsize \sin \theta =\displaystyle \frac{y}{r}$ to find the argument.

Step 3: Write the solution

\scriptsize \begin{align*}z & =r(\cos \theta +i\sin \theta )\\ & =4\sqrt{2}(\cos {{45}^\circ}+i\sin {{45}^\circ})\\ & =4\sqrt{2}cis{{45}^\circ}\end{align*}

### Take note!

When finding the argument, remember that you need to pay attention to which quadrant the complex number is in. A useful strategy can be to make the ratio for $\scriptsize \cos \theta =\displaystyle \frac{x}{r}$ (or $\scriptsize \sin \theta =\displaystyle \frac{y}{r}$) positive, in order to find the reference acute angle $\scriptsize \alpha$, and then to transfer this angle into the necessary quadrant as follows:

• Second quadrant: $\scriptsize {{180}^\circ}-\alpha$
• Third quadrant: $\scriptsize {{180}^\circ}+\alpha$
• Fourth quadrant: $\scriptsize {{360}^\circ}-\alpha$

### Example 3.2

Find the polar form of $\scriptsize z=-4-4i$.

Solution

Step 1: Find the value of $\scriptsize r$ (or $\scriptsize \left| z \right|$ or the modulus)

\scriptsize \begin{align*}r & =\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\ & =\sqrt{{{{{(-4)}}^{2}}+{{{(-4)}}^{2}}}}\\ & =\sqrt{{16+16}}\\ & =\sqrt{{32}}\\ & =4\sqrt{2}\end{align*}

Step 2: Find the value of $\scriptsize \theta$ (the argument)

$\scriptsize z$ is in the third quadrant. Find the reference angle $\scriptsize \alpha$.
$\scriptsize \cos \alpha =\displaystyle \frac{x}{r}=\displaystyle \frac{4}{{4\sqrt{2}}}=\displaystyle \frac{1}{{\sqrt{2}}}$

While we can use the special angle of $\scriptsize {{45}^\circ}$, we need to recognise that the point lies in the third quadrant where the ratios for cosine and sine are negative. Both the $\scriptsize x$ (or real) and $\scriptsize y$ (or imaginary) components are negative.

So while the reference angle is $\scriptsize {{45}^\circ}$, we need to move this into the third quadrant by adding $\scriptsize {{180}^\circ}$ to it. Therefore $\scriptsize \theta ={{180}^\circ}+{{45}^\circ}={{225}^\circ}$.

Step 3: Write the solution

\scriptsize \begin{align*}z & =r(\cos \theta +i\sin \theta )\\ & =4\sqrt{2}(\cos {{225}^\circ}+i\sin {{225}^\circ})\\ & =4\sqrt{2}\text{cis22}{{5}^\circ}\end{align*}

Remember that this is just a shorthand expression for the full polar form of $\scriptsize z=4\sqrt{2}(cos{{225}^\circ}+i\sin {{225}^\circ})$.

### Exercise 3.1

Write the following complex numbers in polar form:

1. $\scriptsize z=5-12i$
2. $\scriptsize z=-7+2i$

The full solutions are at the end of the unit.

## Convert polar form to rectangular form

Sometimes it is necessary to convert a complex number in polar form into standard or rectangular form. To do this, we need to first evaluate the trigonometric functions $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$, then multiply through by $\scriptsize r$.

### Example 3.3

Convert $\scriptsize z=12\text{cis}{{30}^\circ}$ into standard/rectangular form.

Solution

Step 1: Evaluate $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$

We can use the fact that $\scriptsize {{30}^\circ}$ is a special angle to evaluate without a calculator.

$\scriptsize \cos {{30}^\circ}=\displaystyle \frac{{\sqrt{3}}}{2}$
$\scriptsize \sin {{30}^\circ}=\displaystyle \frac{1}{2}$

Step 2: Find the standard form

We know that the complex number is in the first quadrant where cosine and sine are both positive.

\scriptsize \begin{align*}z & =12(\cos {{30}^\circ}+i\sin {{30}^\circ})\\ & =12\left( {\displaystyle \frac{{\sqrt{3}}}{2}+\displaystyle \frac{1}{2}i} \right)\\ & =\displaystyle \frac{{12\sqrt{3}}}{2}+6i\\ & =6\sqrt{3}+6i\end{align*}

The complex number in standard form is $\scriptsize z=6\sqrt{3}+6i$.

### Example 3.4

Convert $\scriptsize z=4\text{cis12}{{\text{0}}^\circ}$ into standard/rectangular form.

Solution

Step 1: Evaluate $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$

$\scriptsize {{120}^\circ}$ is not a special angle but $\scriptsize {{180}^\circ}-{{120}^\circ}={{60}^\circ}$ is a special angle. Therefore, we can evaluate $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$ without a calculator and transfer the angle into the correct quadrant, in this case, the second quadrant.

$\scriptsize \cos {{60}^\circ}=\displaystyle \frac{1}{2}$
$\scriptsize \sin {{60}^\circ}=\displaystyle \frac{{\sqrt{3}}}{2}$

Step 2: Find the standard form

We know that the complex number is in the second quadrant where cosine is negative and sine is positive.

\scriptsize \begin{align*}z &=4(\cos {{120}^\circ}+i\sin {{120}^\circ})\\ & =4\left( {-\displaystyle \frac{1}{2}+\displaystyle \frac{{\sqrt{3}}}{2}i} \right) && \text{Remember that cosine is negative in the }\\ &&&\text{second quadrant}\\ &=-2+\displaystyle \frac{{4\sqrt{3}}}{2}i\\ &=-2+2\sqrt{3}i\end{align*}

The complex number in standard form is $\scriptsize z=-2+2\sqrt{3}i$.

### Example 3.5

Convert $\scriptsize z=3\text{cis4}{{\text{3}}^\circ}$ into standard/rectangular form.

Solution

Step 1: Evaluate $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$

$\scriptsize {{43}^\circ}$ is not a special angle. Therefore, we have to evaluate $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$ with a calculator. The complex number is in the first quadrant.

$\scriptsize \cos {{43}^\circ}=0.7314$
$\scriptsize \sin {{43}^\circ}=0.6820$

Step 2: Find the standard form

We know that the complex number is in the first quadrant where cosine and sine are both positive.

\scriptsize \displaystyle \begin{align*}z & =3(\cos {{43}^\circ}+i\sin {{43}^\circ})\\ & =3\left( {0.7314+0.6820i} \right)\\ & =2.9142+2.046i\end{align*}

The complex number in standard form is $\scriptsize z=2.9142+2.046i$.

### Exercise 3.2

Convert the following complex numbers into standard/rectangular form:

1. $\scriptsize z=6\text{cis}{{45}^\circ}$
2. $\scriptsize z=\sqrt{5}\text{cis21}{{\text{0}}^\circ}$
3. $\scriptsize z=2\text{cis}{{40}^\circ}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to represent complex numbers in polar form.
• How to convert between standard/rectangular form and polar form.

# Unit 3: Assessment

#### Suggested time to complete: 15 minutes

1. Represent the following in polar form:
1. $\scriptsize z=\sqrt{3}+1i$
2. $\scriptsize z=8-4i$
2. Represent the following in rectangular form:
1. $\scriptsize z=3\text{cis}{{240}^\circ}$
2. $\scriptsize z=7\text{cis}{{25}^\circ}$

The full solutions are at the end of the unit.

# Unit 3: Solutions

### Exercise 3.1

1. $\scriptsize z=5-12i$
Find $\scriptsize r$:
\scriptsize \begin{align*}r & =\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\ & =\sqrt{{{{5}^{2}}+{{{12}}^{2}}}}\\ & =\sqrt{{25+144}}\\ & =\sqrt{{169}}\\ & =13\end{align*}
Find $\scriptsize \theta$ ($\scriptsize \theta$ is in the fourth quadrant):
\scriptsize \begin{align*}\cos \theta & =\displaystyle \frac{x}{r}=\displaystyle \frac{5}{{13}}\\\therefore \text{reference angle} & ={{67.38}^\circ}\\\therefore \theta & ={{360}^\circ}-{{67.38}^\circ}={{292.62}^\circ}\end{align*}
Polar form: $\scriptsize z=13\text{cis}{{292.62}^\circ}$
2. $\scriptsize z=-7+2i$
Find $\scriptsize r$:
\scriptsize \begin{align*}r & =\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\ & =\sqrt{{{{{(-7)}}^{2}}+{{2}^{2}}}}\\& =\sqrt{{49+4}}\\ & =\sqrt{{53}}\end{align*}
Find $\scriptsize \theta$ ($\scriptsize \theta$ is in the second quadrant):
$\scriptsize \cos \theta =\displaystyle \frac{x}{r}=\displaystyle \frac{{-7}}{{\sqrt{{53}}}}$
Work with $\scriptsize \cos \theta =\displaystyle \frac{7}{{\sqrt{{53}}}}$ to find a referance angle.
\scriptsize \begin{align*}\therefore \text{reference angle} & ={{15.95}^\circ}\\\therefore \theta & ={{180}^\circ}-{{15.95}^\circ}={{164.05}^\circ}\end{align*}
Polar form: $\scriptsize z=\sqrt{{53}}\text{cis164}\text{.0}{{\text{5}}^\circ}$

Back to Exercise 3.1

### Exercise 3.2

1. $\scriptsize z=6\text{cis}{{45}^\circ}$
$\scriptsize {{45}^\circ}$ is a special angle and the complex number is in the first quadrant.
$\scriptsize \cos {{45}^\circ}=\displaystyle \frac{1}{{\sqrt{2}}}$
$\scriptsize \sin {{45}^\circ}=\displaystyle \frac{1}{{\sqrt{2}}}$
\scriptsize \displaystyle \begin{align*}z & =6(\cos {{45}^\circ}+i\sin {{45}^\circ})\\ & =6\left( {\displaystyle \frac{1}{{\sqrt{2}}}+\displaystyle \frac{1}{{\sqrt{2}}}i} \right)\\ & =\displaystyle \frac{6}{{\sqrt{2}}}+\displaystyle \frac{6}{{\sqrt{2}}}i\end{align*}
The complex number in standard form is $\scriptsize \displaystyle z=\displaystyle \frac{6}{{\sqrt{2}}}+\displaystyle \frac{6}{{\sqrt{2}}}i$.
2. $\scriptsize z=\sqrt{5}\text{cis21}{{\text{0}}^\circ}$
$\scriptsize {{180}^\circ}+{{30}^\circ}={{210}^\circ}$. Therefore, we use the special angle of $\scriptsize {{30}^\circ}$ in the third quadrant.
$\scriptsize \cos {{210}^\circ}=\cos ({{180}^\circ}+{{30}^\circ})=-\cos {{30}^\circ}=-\displaystyle \frac{{\sqrt{3}}}{2}$
$\scriptsize \sin {{210}^\circ}=\sin ({{180}^\circ}+{{30}^\circ})=-\sin {{30}^\circ}=-\displaystyle \frac{1}{2}$
\scriptsize \displaystyle \begin{align*}z & =\sqrt{5}(\cos {{210}^\circ}+i\sin {{210}^\circ})\\ & =\sqrt{5}\left( {-\displaystyle \frac{{\sqrt{3}}}{2}-\displaystyle \frac{1}{2}i} \right)\quad \text{Both cosine and sine are negative in the third quadrant}\\ & =-\displaystyle \frac{{\sqrt{{15}}}}{2}-\displaystyle \frac{{\sqrt{5}}}{2}i\end{align*}
The complex number in standard form is $\scriptsize \displaystyle z=-\displaystyle \frac{{\sqrt{{15}}}}{2}-\displaystyle \frac{{\sqrt{5}}}{2}i$.
3. $\scriptsize z=2\text{cis}{{40}^\circ}$
$\scriptsize {{40}^\circ}$ is not a special angle. The complex number is in the first quadrant.
$\scriptsize \cos {{40}^\circ}=0.7660$
$\scriptsize \sin {{40}^\circ}=0.6428$
\scriptsize \displaystyle \begin{align*}z & =2(\cos {{40}^\circ}+i\sin {{40}^\circ})\\ & =2\left( {0.7660+0.6428i} \right)\quad \text{Both cosine and sine are positive in the first quadrant}\\ & =1.532+1.2856i\end{align*}
The complex number in standard form is $\scriptsize \displaystyle z=1.532+1.2856i$.

Back to Exercise 3.2

### Unit 3: Assessment

1. .
1. $\scriptsize z=\sqrt{3}+1i$
Find $\scriptsize r$:
\scriptsize \begin{align*}r & =\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\ & =\sqrt{{{{{\left( {\sqrt{3}} \right)}}^{2}}+{{1}^{2}}}}\\ & =\sqrt{{3+1}}\\ & =2\end{align*}
Find $\scriptsize \theta$:
The complex number is in the first quadrant.
\scriptsize \begin{align*}\cos \theta & =\displaystyle \frac{x}{r}=\displaystyle \frac{{\sqrt{3}}}{2}\\\therefore \theta & ={{30}^\circ}\end{align*}
Polar form: $\scriptsize z=2\text{cis}{{30}^\circ}$
2. $\scriptsize z=8-4i$
Find $\scriptsize r$:
\scriptsize \begin{align*}r & =\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\ & =\sqrt{{{{8}^{2}}+{{{(-4)}}^{2}}}}\\ & =\sqrt{{64+16}}\\ & =\sqrt{{80}}\\ & =4\sqrt{5}\end{align*}
Find $\scriptsize \theta$:
The complex number is in the fourth quadrant.
\scriptsize \begin{align*}\cos \theta & =\displaystyle \frac{x}{r}=\displaystyle \frac{8}{{4\sqrt{5}}}=\displaystyle \frac{2}{{\sqrt{5}}}\\\therefore \text{Reference angle} & ={{26.57}^\circ}\\\therefore \theta & ={{360}^\circ}-{{26.57}^\circ}={{333.43}^\circ}\end{align*}
Polar form: $\scriptsize z=4\sqrt{5}\text{cis}{{333.43}^\circ}$
2. .
1. $\scriptsize z=3\text{cis}{{240}^\circ}$
$\scriptsize {{240}^\circ}$ is not a special angle but $\scriptsize {{240}^\circ}-{{180}^\circ}={{60}^\circ}$ is a special angle. Therefore, we can evaluate $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$ without a calculator and transfer the angle into the correct quadrant, in this case, the third quadrant.
$\scriptsize \cos {{60}^\circ}=\displaystyle \frac{1}{2}$
$\scriptsize \sin {{60}^\circ}=\displaystyle \frac{{\sqrt{3}}}{2}$
The complex number is in the third quadrant where cosine and sine are both negative.
\scriptsize \displaystyle \begin{align*}z & =3(\cos {{240}^\circ}+i\sin {{240}^\circ})\\ & =3\left( {-\displaystyle \frac{1}{2}-\displaystyle \frac{{\sqrt{3}}}{2}i} \right)\quad \text{cosine and sine are negative in the third quadrant}\\ & =-\displaystyle \frac{3}{2}-\displaystyle \frac{{3\sqrt{3}}}{2}i\end{align*}
Standard form is $\scriptsize z=-\displaystyle \frac{3}{2}-\displaystyle \frac{{3\sqrt{3}}}{2}i$.
2. $\scriptsize z=7\text{cis}{{25}^\circ}$
$\scriptsize {{25}^\circ}$ is not a special angle. The complex number is in the first quadrant.
$\scriptsize \cos {{25}^\circ}=0.9063$
$\scriptsize \sin {{25}^\circ}=0.4226$The complex number is in the first quadrant where cosine and sine are positive.
\scriptsize \displaystyle \begin{align*}z & =7(\cos {{25}^\circ}+i\sin {{25}^\circ})\\ & =7\left( {0.9063+0.4226i} \right)\quad \text{cosine and sine are positive in the first quadrant}\\ & =6.3441+2.9582i\end{align*}
Standard form is $\scriptsize z=6.3441+2.9582i$.

Back to Unit 3: Assessment