Complex Numbers: Working with complex numbers

# Unit 2: Revise the polar form of complex numbers

Dylan Busa

### Unit outcomes

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

• Plot a complex number on the complex plan.
• Find the absolute value of a complex number.
• Convert a complex number from standard (or rectangular) form to polar form.
• Convert a complex number from polar form to standard (or rectangular) form.
• Understand what is meant by the abbreviation when dealing with complex numbers in polar

## What you should know

Before you start this unit, make sure you can:

## Introduction

This unit revises the polar form of complex numbers covered in level 3 subject outcome 1.1 units 2 and 3. It is important that you complete these subject outcomes before continuing.

To understand what the polar form of a complex number is and where it comes from, we need to understand how to plot complex numbers on the complex plane.

## The complex plane

We know that we can plot the position of any real number on a number line as shown in Figure 1.

But complex numbers have a real and an imaginary part, therefore we need two numbers to plot them – one for the real part (the x-axis) and one for the imaginary part (the y-axis) placed at right angles to each other to create the complex plane. This is a coordinate system like the Cartesian plane and complex numbers are points on the plane, expressed as ordered pairs.

Take the complex number $\scriptsize 2-3i$, for example. We can plot it on the complex plane as shown in Figure 2. It is the point $\scriptsize (2,-3)$.

### Exercise 2.1

Plot the following complex numbers on the same complex plane:

1. $\scriptsize -4-5i$
2. $\scriptsize 4+3i$
3. $\scriptsize -3$
4. $\scriptsize -4i$

The full solutions are at the end of the unit.

### Take note!

An Argand diagram is a plot of complex numbers as points on the complex plane using the x-axis as the real axis and y-axis as the imaginary axis.

## The modulus and argument

We often refer to complex numbers as $\scriptsize z$ and the complex plane as the z-plane. We can say that $\scriptsize z=3+4i$.

Figure 3 shows the complex number $\scriptsize z=3+4i$. The distance of this point from the origin is called the modulus and is designated as $\scriptsize \left| z \right|$.

We can determine the modulus by dropping a perpendicular from this point (see Figure 4) and using Pythagoras’ theorem.

\scriptsize \begin{align*}{{z}^{2}} & ={{3}^{3}}+{{4}^{2}}\\ & =9+16=25\\\therefore \left| z \right| & =\sqrt{{25}}=5\end{align*}

The modulus is always represented as an absolute value with $\scriptsize \left| {} \right|$ signs because it is a length and so always taken as the postive value. In the calculation of $\scriptsize \left| z \right|$ above, we ignore $\scriptsize -5$ as a possible solution.

### Example 2.1

Given $\scriptsize z=5-6i$, find $\scriptsize \left| z \right|$.

Solution

We can plot the complex number $\scriptsize z=5-6i$ on the complex plane.

\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{x}^{2}}+{{y}^{2}}}}\quad x\text{ is the real part and }y\text{ is the imaginary part}\\&=\sqrt{{{{5}^{2}}+{{{(-6)}}^{2}}}}\\&=\sqrt{{25+36}}\\&=\sqrt{{61}}\end{align*}

### Exercise 2.2

Find $\scriptsize \left| z \right|$ in each case:

1. $\scriptsize z=1+7i$
2. $\scriptsize z=-3-5i$
3. $\scriptsize z=-4+\displaystyle \frac{3}{2}i$
4. $\scriptsize z=-\sqrt{5}-\sqrt{{-6}}$

The full solutions are at the end of the unit.

The modulus is not enough to fully define the position of a complex number on the complex plane. Figure 5 shows two complex numbers, $\scriptsize {{z}_{1}}=4+3i$ and $\scriptsize {{z}_{2}}=3-4i$. For each, $\scriptsize \left| z \right|=5$.

We also need to know the argument, the angle the line representing the modulus makes with the positive x-axis (see Figure 6).

In Figure 6 we see that $\scriptsize \sin \alpha =\displaystyle \frac{y}{r}=\displaystyle \frac{3}{5}$ or that $\scriptsize \cos \alpha =\displaystyle \frac{x}{r}=\displaystyle \frac{4}{5}$. Therefore, $\scriptsize \alpha ={{36.870}^\circ}$. This answer makes sense because $\scriptsize {{z}_{1}}=4+3i$ is in the first quadrant.

Also, we can say that $\scriptsize \sin \beta =\displaystyle \frac{y}{r}=\displaystyle \frac{{-4}}{5}$ or that $\scriptsize \cos \beta =\displaystyle \frac{x}{r}=\displaystyle \frac{3}{5}$. Therefore, $\scriptsize \beta =-{{53.130}^\circ}$. This makes sense because $\scriptsize {{z}_{2}}=3-4i$ is in the fourth quadrant.

With the modulus and the argument, we can uniquely specify the position of any complex number on the complex plane.

### Take note!

The value of the modulus $\scriptsize \left| z \right|$ is the same as the value of $\scriptsize r$ in $\scriptsize \sin \theta =\displaystyle \frac{y}{r}$ or $\scriptsize \cos \theta =\displaystyle \frac{x}{r}$.

### Take note!

When calculating the argument, it is important that you draw a sketch of the position of $\scriptsize z$ on the complex plane to determine which quadrant the point is in. The three trigonometric ratios still follow the CAST diagram on the complex plane (see Figure 7).

### Example 2.2

Determine the modulus and the argument of $\scriptsize z=-2-\sqrt{{-3}}$.

Solution

First, write the complex number in standard form $\scriptsize a+bi$.
$\scriptsize z=-2-\sqrt{{-3}}=-2-\sqrt{3}i$

$\scriptsize z$ lies in the third quadrant because $\scriptsize x$ ($\scriptsize -2$) and $\scriptsize y$ ($\scriptsize -\sqrt{3}$) are both negative.

\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{x}^{2}}+{{y}^{2}}}}\\&=\sqrt{{{{{(-2)}}^{2}}+{{{\left( {-\sqrt{3}} \right)}}^{2}}}}\\&=\sqrt{{4+3}}\\&=\sqrt{7}\end{align*}

To find the argument, start by finding a reference angle ($\scriptsize \alpha$) using a positive ratio of sine or cosine.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{y}{r}=\displaystyle \frac{{\sqrt{3}}}{{\sqrt{7}}}\\\therefore \alpha & ={{40.89}^\circ}\end{align*}

But $\scriptsize z$ lies in the third quadrant. Therefore, $\scriptsize \theta ={{180}^\circ}+{{40.89}^\circ}={{220.89}^\circ}$.

### 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 $\scriptsize \sin \theta =\displaystyle \frac{y}{r}$ or $\scriptsize \cos \theta =\displaystyle \frac{x}{r}$ positive initially, in order to find the acute reference angle $\scriptsize \theta$, and then to transfer this angle into the necessary quadrant as indicated in Figure 8. To transfer these angles, use the following identities:

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

### Exercise 2.3

Determine the modulus and the argument of the following complex numbers, leaving your moduli in surd form:

1. $\scriptsize z=-1-8i$
2. $\scriptsize z=-2+2i$
3. $\scriptsize z=4-3i$

The full solutions are at the end of the unit.

## 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 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 (see Figure 9).

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 \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 2.3

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

Solution

Step 1: Determine $\scriptsize r$ (or $\scriptsize \left| z \right|$)
\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: Determine $\scriptsize \theta$
$\scriptsize z$ is in the third quadrant. Find reference angle $\scriptsize \alpha$.
\scriptsize \begin{align*}\cos \alpha & =\displaystyle \frac{x}{r}=\displaystyle \frac{4}{{4\sqrt{2}}}=\displaystyle \frac{1}{{\sqrt{2}}}\\\therefore \alpha & ={{45}^\circ}\end{align*}
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*}

### Exercise 2.4

Write the following complex numbers in polar form:

1. $\scriptsize z=5+12i$
2. $\scriptsize z=-7-3i$
3. $\scriptsize z=\sqrt{5}-\sqrt{{-6}}$

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 2.4

Convert $\scriptsize z=12\text{cis4}{{\text{5}}^\circ}$ into standard/rectangular form.

Solution

Step 1: Evaluate $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$
We can use the fact that $\scriptsize {{45}^\circ}$ is a special angle to evaluate without a calculator.
$\scriptsize \cos {{45}^\circ}=\displaystyle \frac{1}{{\sqrt{2}}}$
$\scriptsize \sin {{45}^\circ}=\displaystyle \frac{1}{{\sqrt{2}}}$

Step 2: Find the standard form
$\scriptsize z$ is in the first quadrant where cosine and sine are positive.
\scriptsize \begin{align*}z&=12(\cos {{45}^\circ}+i\sin {{45}^\circ})\\&=12\left( {\displaystyle \frac{1}{{\sqrt{2}}}+\displaystyle \frac{1}{{\sqrt{2}}}i} \right)\\&=\displaystyle \frac{{12}}{{\sqrt{2}}}+\displaystyle \frac{{12}}{{\sqrt{2}}}i\\&=\displaystyle \frac{{12\sqrt{2}}}{2}+\displaystyle \frac{{12\sqrt{2}}}{2}i\\&=6\sqrt{2}+6\sqrt{2}i\end{align*}

The complex number in standard form is $\scriptsize z=6\sqrt{2}+6\sqrt{2}i$.

### Example 2.5

Convert $\scriptsize z=4\text{cis15}{{\text{0}}^\circ}$ into standard/rectangular form.

Solution

Step 1: Evaluate $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$
$\scriptsize {{120}^\circ}$ is not a special angle but $\scriptsize {{180}^\circ}-{{150}^\circ}={{30}^\circ}$ 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.
$\scriptsize \cos {{30}^\circ}=\displaystyle \frac{{\sqrt{3}}}{2}$
$\scriptsize \sin {{30}^\circ}=\displaystyle \frac{1}{2}$

Step 2: Find the standard form
We know that the complex number is in the second quadrant where cosine is negative and sine is positive.
\scriptsize \begin{align*}z&=4(\cos {{150}^\circ}+i\sin {{150}^\circ})\\&=4\left( {-\displaystyle \frac{{\sqrt{3}}}{2}+\displaystyle \frac{1}{2}i} \right)\quad \text{Remember that cosine is negative in the second quadrant}\\&=-2\sqrt{3}+2i\end{align*}

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

### Example 2.6

Convert $\scriptsize z=3\text{cis24}{{\text{3}}^\circ}$ into standard/rectangular form.

Solution

Step 1: Evaluate $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$
$\scriptsize {{243}^\circ}$ is not a special angle and cannot be reduced to a special angle. Therefore, we have to evaluate $\scriptsize \cos \theta$ and $\scriptsize \sin \theta$ with a calculator. The complex number is in the third quadrant.
$\scriptsize \cos {{243}^\circ}=-0.454$
$\scriptsize \sin {{243}^\circ}=-0.891$

Step 2: Find the standard form
We know that the complex number is in the third quadrant where cosine and sine are negative.
\scriptsize \begin{align*}z&=3(\cos {{243}^\circ}+i\sin {{243}^\circ})\\&=3\left( {-0.454-0.891i} \right)\\&=-1.362-2.673i\end{align*}

The complex number in standard form is $\scriptsize z=-1.362-2.673i$.

### Exercise 2.5

Convert the following complex numbers into standard/rectangular form:

1. $\scriptsize z=\sqrt{3}\text{cis}{{45}^\circ}$
2. $\scriptsize z=\sqrt{7}\text{cis21}{{\text{0}}^\circ}$
3. $\scriptsize z=3\text{cis3}{{40}^\circ}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to plot a complex number on the complex plane.
• How to find the absolute value or modulus of a complex number.
• How to convert a complex number from standard (or rectangular) form to polar form.
• How to convert a complex number from polar form to standard (or rectangular) form.
• How to use the $\scriptsize \text{cis}$ shorthand for the polar form.

# Unit 2: Assessment

#### Suggested time to complete: 25 minutes

1. Given $\scriptsize z=-1+\sqrt{3}i$:
1. Write down the conjugate of $\scriptsize z$.
2. Calculate the modulus of $\scriptsize z$.
3. Determine the argument of $\scriptsize z$.
4. Write $\scriptsize z$ in polar form.
2. Write the following complex numbers in polar form:
1. $\scriptsize z=\displaystyle \frac{3}{2}-7i$
2. $\scriptsize z=-4+\sqrt{{-8}}$
3. $\scriptsize z=\sqrt{{13}}-\sqrt{{-13}}$
4. $\scriptsize z=-6+3.543i$
3. Write the following complex numbers in standard form:
1. $\scriptsize z=13\text{cis}{{330}^\circ}$
2. $\scriptsize z=\sqrt{7}\text{cis22}{{\text{5}}^\circ}$
3. $\scriptsize z=\displaystyle \frac{2}{{\sqrt{3}}}\text{cis12}{{\text{0}}^\circ}$
4. $\scriptsize z=4\text{cis16}{{\text{0}}^\circ}$

The full solutions are at the end of the unit.

# Unit 2: Solutions

### Exercise 2.1

Back to Exercise 2.1

### Exercise 2.2

1. .
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{1}^{2}}+{{7}^{2}}}}\\&=\sqrt{{1+49}}\\&=\sqrt{{50}}\\&=5\sqrt{2}\end{align*}
2. .
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{(-3)}}^{2}}+{{{(-5)}}^{2}}}}\text{ }\\&=\sqrt{{9+25}}\\&=\sqrt{{34}}\end{align*}
3. .
\scriptsize \begin{align*}\mid z\mid&=\sqrt{(-4)^2+\left(\displaystyle \frac{3}{2}\right)^2}\\ &=\sqrt{16+\displaystyle \frac{9}{4}}\\ &=\sqrt{\displaystyle \frac{64+9}{4}}\\ &=\sqrt{\displaystyle \frac{73}{4}}\\ &=\displaystyle \frac{\sqrt{73}}{2}\end{align*}
4. .
$\scriptsize z=-\sqrt{5}-\sqrt{{-6}}=-\sqrt{5}-\sqrt{6}i$
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{\left( {-\sqrt{5}} \right)}}^{2}}+{{{\left( {-\sqrt{6}} \right)}}^{2}}}}\\&=\sqrt{{5+6}}\\&=\sqrt{{11}}\end{align*}

Back to Exercise 2.2

### Exercise 2.3

1. $\scriptsize z=-1-8i$
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{(-1)}}^{2}}+{{{(-8)}}^{2}}}}\\&=\sqrt{{1+64}}\\&=\sqrt{{65}}\end{align*}
$\scriptsize z$ is in the third quadrant. Find reference angle $\scriptsize \alpha$.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{8}{{\sqrt{{65}}}}\\\therefore \alpha & ={{82.87}^\circ}\end{align*}
Therefore, $\scriptsize \theta ={{180}^\circ}+\alpha ={{262.87}^\circ}$.
2. $\scriptsize z=-2+2i$
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{(-2)}}^{2}}+{{2}^{2}}}}\\&=\sqrt{{4+4}}\\&=\sqrt{8}\\&=2\sqrt{2}\end{align*}
$\scriptsize z$ is in the second quadrant. Find reference angle $\scriptsize \alpha$.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{2}{{2\sqrt{2}}}=\displaystyle \frac{1}{{\sqrt{2}}}\\\therefore \alpha & ={{45}^\circ}\end{align*}
Therefore, $\scriptsize \theta ={{180}^\circ}-\alpha ={{135}^\circ}$.
3. $\scriptsize z=4-3i$
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{(4)}}^{2}}+{{{(-3)}}^{2}}}}\\&=\sqrt{{16+9}}\\&=\sqrt{{25}}\\&=5\end{align*}
$\scriptsize z$ is in the fourth quadrant. Find reference angle $\scriptsize \alpha$.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{3}{5}\\\therefore \alpha & ={{36.87}^\circ}\end{align*}
Therefore, $\scriptsize \theta ={{360}^\circ}-\alpha ={{323.13}^\circ}$.

Back to Exercise 2.3

### Exercise 2.4

1. $\scriptsize z=5+12i$
\scriptsize \begin{align*}r&=\sqrt{{{{{(5)}}^{2}}+{{{(12)}}^{2}}}}\\&=\sqrt{{25+144}}\\&=\sqrt{{169}}\\&=13\end{align*}
$\scriptsize z$ is in the first quadrant.
\scriptsize \begin{align*}\sin \theta & =\displaystyle \frac{{12}}{{13}}\\\therefore \theta & ={{67.38}^\circ}\end{align*}
\scriptsize \begin{align*}z&=r(\cos \theta +i\sin \theta )\\&=13(\cos {{67.38}^\circ}+i\sin {{67.38}^\circ})\\&=13\text{cis}{{67.38}^\circ}\end{align*}
2. $\scriptsize z=-7-3i$
\scriptsize \begin{align*}r&=\sqrt{{{{{(-7)}}^{2}}+{{{(-3)}}^{2}}}}\\&=\sqrt{{49+9}}\\&=\sqrt{{58}}\end{align*}
$\scriptsize z$ is in the third quadrant.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{3}{{\sqrt{{58}}}}\\\therefore \alpha &={{23.20}^\circ}\end{align*}
Therefore, $\scriptsize \theta ={{180}^\circ}+{{23.20}^\circ}={{203.20}^\circ}$
\scriptsize \begin{align*}z&=r(\cos \theta +i\sin \theta )\\&=\sqrt{{58}}(\cos {{203.30}^\circ}+i\sin {{203.20}^\circ})\\&=\sqrt{{58}}\text{cis}{{203.20}^\circ}\end{align*}
3. $\scriptsize z=\sqrt{5}-\sqrt{{-6}}=\sqrt{5}-\sqrt{6}i$
\scriptsize \begin{align*}r&=\sqrt{{{{{\left( {\sqrt{5}} \right)}}^{2}}+{{{\left( {\sqrt{6}} \right)}}^{2}}}}\\&=\sqrt{{5+6}}\\&=\sqrt{{11}}\end{align*}
$\scriptsize z$ is in the fourth quadrant.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{{\sqrt{6}}}{{\sqrt{{11}}}}\\\therefore \alpha &={{47.61}^\circ}\end{align*}
Therefore, $\scriptsize \theta ={{360}^\circ}-{{47.61}^\circ}={{312.39}^\circ}$.
\scriptsize \begin{align*}z&=r(\cos \theta +i\sin \theta )\\&=\sqrt{{11}}(\cos {{312.39}^\circ}+i\sin {{312.39}^\circ})\\&=\sqrt{{11}}\text{cis}{{312.39}^\circ}\end{align*}

Back to Exercise 2.4

### Exercise 2.5

1. $\scriptsize z=\sqrt{3}\text{cis}{{45}^\circ}$
$\scriptsize \cos {{45}^\circ}=\displaystyle \frac{1}{{\sqrt{2}}}$
$\scriptsize \sin {{45}^\circ}=\displaystyle \frac{1}{{\sqrt{2}}}$
\scriptsize \begin{align*}z&=\sqrt{3}\left( {\displaystyle \frac{1}{{\sqrt{2}}}+\displaystyle \frac{1}{{\sqrt{2}}}i} \right)\\&=\displaystyle \frac{{\sqrt{3}}}{{\sqrt{2}}}+\displaystyle \frac{{\sqrt{3}}}{{\sqrt{2}}}i\\&=\displaystyle \frac{{\sqrt{2}\sqrt{3}}}{2}+\displaystyle \frac{{\sqrt{2}\sqrt{3}}}{2}i\end{align*}
2. $\scriptsize z=\sqrt{7}\text{cis21}{{\text{0}}^\circ}$
$\scriptsize \cos {{210}^\circ}=\cos ({{180}^\circ}+{{30}^\circ})=-\cos {{30}^\circ}=-\displaystyle \frac{{\sqrt{3}}}{2}$
$\scriptsize \sin {{210}^\circ}=\sin ({{180}^\circ}+{{30}^\circ})=-\sin {{30}^\circ}=-\displaystyle \frac{1}{2}$
\scriptsize \begin{align*}z&=\sqrt{7}\left( {-\displaystyle \frac{{\sqrt{3}}}{2}-\displaystyle \frac{1}{2}i} \right)\\&=\displaystyle \frac{{\sqrt{{21}}}}{2}-\displaystyle \frac{{\sqrt{7}}}{2}i\end{align*}
3. $\scriptsize z=3\text{cis3}{{40}^\circ}$
$\scriptsize \cos {{340}^\circ}=0.940$
$\scriptsize \sin {{340}^\circ}=-0.342$
\scriptsize \begin{align*}z&=3\left( {0.940-0.342i} \right)\\&=2.82-1.026i\end{align*}

Back to Exercise 2.5

### Unit 2: Assessment

1. $\scriptsize z=-1+\sqrt{3}i$
1. $\scriptsize -1-\sqrt{3}i$
2. .
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{\left( {-1} \right)}}^{2}}+{{{\left( {\sqrt{3}} \right)}}^{2}}}}\\&=\sqrt{{1+3}}\\&=\sqrt{4}\\&=2\end{align*}
3. $\scriptsize z$ is in the second quadrant.
\scriptsize \begin{align*}\sin \alpha &=\displaystyle \frac{{\sqrt{3}}}{2}\\\therefore \alpha &={{60}^\circ}\end{align*}
$\scriptsize \theta ={{180}^\circ}-{{60}^\circ}={{120}^\circ}$
4. .
EQU
2. .
1. $\scriptsize z=\displaystyle \frac{3}{2}-7i$
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{\left( {\displaystyle \frac{3}{2}} \right)}}^{2}}+{{{\left( {-7} \right)}}^{2}}}}\\&=\sqrt{{\displaystyle \frac{9}{4}+49}}\\&=\sqrt{{\displaystyle \frac{{9+196}}{4}}}\\&=\sqrt{{\displaystyle \frac{{205}}{4}}}\\&=\displaystyle \frac{{\sqrt{{205}}}}{2}\end{align*}
$\scriptsize z$ is in the fourth quadrant.
\scriptsize \begin{align*}\sin \alpha &=\displaystyle \frac{7}{{\displaystyle \frac{{\sqrt{{205}}}}{2}}}\\\therefore \alpha &={{77.91}^\circ}\end{align*}
$\scriptsize \theta ={{360}^\circ}-{{77.91}^\circ}={{282.09}^\circ}$
\scriptsize \begin{align*}z&=\displaystyle \frac{{\sqrt{{205}}}}{2}(\cos {{282.09}^\circ}+i\sin {{282.09}^\circ})\\&=\displaystyle \frac{{\sqrt{{205}}}}{2}\text{cis}{{282.09}^\circ}\end{align*}
2. $\scriptsize z=-4+\sqrt{{-8}}=-4+\sqrt{8}i$
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{(-4)}}^{2}}+{{8}^{2}}}}\\&=\sqrt{{16+64}}\\&=\sqrt{{80}}\\&=4\sqrt{5}\end{align*}
$\scriptsize z$ is in the second quadrant.
\scriptsize \begin{align*}\sin \alpha &=\displaystyle \frac{{\sqrt{8}}}{{4\sqrt{5}}}\\\therefore \alpha &={{18.43}^\circ}\end{align*}
$\scriptsize \theta ={{180}^\circ}-{{18.43}^\circ}={{161.57}^\circ}$
\scriptsize \begin{align*}z&=4\sqrt{5}(\cos {{161.57}^\circ}+i\sin {{161.57}^\circ})\\&=4\sqrt{5}\text{cis}{{161.57}^\circ}\end{align*}
3. $\scriptsize z=\sqrt{{13}}-\sqrt{{-13}}=\sqrt{{13}}-\sqrt{{13}}i$
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{\left( {\sqrt{{13}}} \right)}}^{2}}+{{{\left( {-\sqrt{{13}}} \right)}}^{2}}}}\\&=\sqrt{{13+13}}\\&=\sqrt{{26}}\end{align*}
$\scriptsize z$ is in the fourth quadrant.
\scriptsize \begin{align*}\sin \alpha &=\displaystyle \frac{{\sqrt{{13}}}}{{\sqrt{{26}}}}=\displaystyle \frac{{\sqrt{{13}}}}{{\sqrt{2}\sqrt{{13}}}}=\displaystyle \frac{1}{{\sqrt{2}}}\\\therefore \alpha &={{45}^\circ}\end{align*}
$\scriptsize \theta ={{360}^\circ}-{{45}^\circ}={{315}^\circ}$
\scriptsize \begin{align*}z&=\sqrt{{26}}(\cos {{315}^\circ}+i\sin {{315}^\circ})\\&=\sqrt{{26}}\text{cis31}{{\text{5}}^\circ}\end{align*}
4. $\scriptsize z=-6+3.543i$
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{(-6)}}^{2}}+{{{3.543}}^{2}}}}\\&=\sqrt{{36+12.55}}\\&=\sqrt{{48.55}}\end{align*}
$\scriptsize z$ is in the second quadrant.
\scriptsize \begin{align*}\sin \alpha &=\displaystyle \frac{{3.543}}{{\sqrt{{48.55}}}}\\\therefore \alpha &={{30.56}^\circ}\end{align*}
$\scriptsize \theta ={{180}^\circ}-{{30.56}^\circ}={{149.44}^\circ}$
\scriptsize \begin{align*}z&=\sqrt{{48.55}}(\cos {{149.44}^\circ}+i\sin {{149.44}^\circ})\\&=\sqrt{{48.55}}\text{cis}{{149.44}^\circ}\end{align*}
3. .
1. $\scriptsize z=13\text{cis}{{330}^\circ}$
$\scriptsize \cos {{330}^\circ}=\cos ({{360}^\circ}-{{30}^\circ})=\cos {{30}^\circ}=\displaystyle \frac{{\sqrt{3}}}{2}$
$\scriptsize \sin {{330}^\circ}=\sin ({{360}^\circ}-{{30}^\circ})=-\sin {{30}^\circ}=-\displaystyle \frac{1}{2}$
\scriptsize \begin{align*}z&=13\left( {\displaystyle \frac{{\sqrt{3}}}{2}-\displaystyle \frac{1}{2}i} \right)\\&=\displaystyle \frac{{13\sqrt{3}}}{2}-\displaystyle \frac{{13}}{2}i\end{align*}
2. $\scriptsize z=\sqrt{7}\text{cis22}{{\text{5}}^\circ}$
$\scriptsize \cos {{225}^\circ}=\cos ({{180}^\circ}+{{45}^\circ})=-\cos {{45}^\circ}=-\displaystyle \frac{1}{{\sqrt{2}}}$
$\scriptsize \sin {{225}^\circ}=\sin ({{180}^\circ}+{{45}^\circ})=-\sin {{45}^\circ}=-\displaystyle \frac{1}{{\sqrt{2}}}$
\scriptsize \begin{align*}z&=\sqrt{7}\left( {-\displaystyle \frac{1}{{\sqrt{2}}}-\displaystyle \frac{1}{{\sqrt{2}}}i} \right)\\&=-\displaystyle \frac{{\sqrt{7}}}{{\sqrt{2}}}-\displaystyle \frac{{\sqrt{7}}}{{\sqrt{2}}}i\end{align*}
3. $\scriptsize z=\displaystyle \frac{2}{{\sqrt{3}}}\text{cis12}{{\text{0}}^\circ}$
$\scriptsize \cos {{120}^\circ}=\cos ({{180}^\circ}-{{60}^\circ})=-\cos {{60}^\circ}=-\displaystyle \frac{1}{2}$
$\scriptsize \sin {{120}^\circ}=\sin ({{180}^\circ}-{{60}^\circ})=\sin {{60}^\circ}=\displaystyle \frac{{\sqrt{3}}}{2}$
\scriptsize \begin{align*}z&=\displaystyle \frac{2}{{\sqrt{3}}}\left( {-\displaystyle \frac{1}{2}+\displaystyle \frac{{\sqrt{3}}}{2}i} \right)\\&=-\displaystyle \frac{1}{{\sqrt{3}}}+i\end{align*}
4. $\scriptsize z=4\text{cis16}{{\text{0}}^\circ}$
$\scriptsize \cos {{160}^\circ}=-0.940$
$\scriptsize \sin {{160}^\circ}=0.342$
\scriptsize \begin{align*}z&=4\left( {-0.940+0.342i} \right)\\&=-3.76+1.368i\end{align*}

Back to Unit 2: Assessment