Complex Numbers: Working with complex numbers

# Unit 4: Raise complex numbers to exponents using De Moivre’s Theorem

Dylan Busa

### Unit outcomes

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

• Find the powers of complex numbers in polar form.
• Simplify complex expressions with powers.

## What you should know

Before you start this unit, make sure you can:

• Multiply complex numbers in polar form. Refer to unit 3 if you need help with this.
• Multiply complex numbers in polar form. Refer to unit 3 if you need help with this.
• Convert between the rectangular form and polar form of complex numbers. Refer to unit 2 if you need help with this.

## Introduction

In this unit we are going to learn how to deal with the powers of complex numbers. As you will see, the polar form is extremely useful in simplifying what would otherwise be very messy and difficult calculations.

## De Moivre’s theorem

Abraham de Moivre (pictured in Figure 1) was a French mathematician of the late 17th and early 18th century. He spent most of his life in England having escaped religious persecution in France, and was a friend of Isaac Newton and James Stirling.

He is most famous for his theorem that links the complex numbers to trigonometry, but perhaps contributed most significantly to the study of probability.

In Activity 4.1, we will discover de Moivre’s theorem for ourselves.

### Activity 4.1: De Moivre’s theorem

Time required: 15 minutes

What you need:

• a pen or pencil
• a blank piece of paper

What to do:

1. If $\scriptsize z=r(\cos \theta +i\sin \theta )$ write down an expression for $\scriptsize {{z}^{2}}$. Hint: What is $\scriptsize {{z}_{1}}{{z}_{2}}$?
2. Using the fact that $\scriptsize {{z}^{3}}={{z}^{2}}\cdot z$, write down an expression for $\scriptsize {{z}^{3}}$.
3. Using the fact that $\scriptsize {{z}^{4}}={{z}^{3}}\cdot z$, write down an expression for $\scriptsize {{z}^{4}}$.
4. Write down an expression for $\scriptsize {{z}^{5}}$.
5. Write down an expression for $\scriptsize {{z}^{n}}$.

What did you find?

1. If $\scriptsize {{z}_{1}}={{r}_{1}}(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})$ and $\scriptsize {{z}_{2}}={{r}_{2}}(\cos {{\theta }_{2}}+i\sin {{\theta }_{2}})$, then we know that $\scriptsize {{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)$. Therefore, we can say that if $\scriptsize z=r(\cos \theta +i\sin \theta )$, then
\scriptsize \begin{align*}{{z}^{2}}&=r\cdot r\left[ {\cos (\theta +\theta )+i\sin (\theta +\theta )} \right]\\&={{r}^{2}}(\cos 2\theta +i\sin 2\theta )\end{align*}
2. $\scriptsize {{z}^{3}} ={{z}^{2}}\cdot z$, therefore:
\scriptsize \begin{align*}{{z}^{3}} & ={{r}^{2}}(\cos 2\theta +i\sin 2\theta )\times r(\cos \theta +i\sin \theta )\\ & ={{r}^{2}}\cdot r\left[ {\cos (2\theta +\theta )+i\sin (2\theta +\theta )} \right]\\ & ={{r}^{3}}(\cos 3\theta +i\sin 3\theta )\end{align*}
3. $\scriptsize {{z}^{4}} ={{z}^{3}}\cdot z$, therefore:
\scriptsize \begin{align*}{{z}^{4}} & ={{r}^{3}}(\cos 3\theta +i\sin 3\theta )\times r(\cos \theta +i\sin \theta )\\ & ={{r}^{3}}\cdot r\left[ {\cos (3\theta +\theta )+i\sin (3\theta +\theta )} \right]\\ & ={{r}^{4}}(\cos 4\theta +i\sin 4\theta )\end{align*}
4. $\scriptsize {{z}^{5}} ={{z}^{4}}\cdot z$, therefore:
\scriptsize \begin{align*}{{z}^{5}} & ={{r}^{4}}(\cos 4\theta +i\sin 4\theta )\times r(\cos \theta +i\sin \theta )\\ & ={{r}^{4}}\cdot r\left[ {\cos (4\theta +\theta )+i\sin (4\theta +\theta )} \right]\\ & ={{r}^{5}}(\cos 5\theta +i\sin 5\theta )\end{align*}
5. $\scriptsize {{z}^{n}}={{r}^{n}}\left[ {\cos (n\theta )+i\sin (n\theta )} \right]$

Congratulations! You have just discovered de Moivre’s theorem for yourself. You are a mathematician! The process we used to derive de Moivre’s theorem is called mathematical induction and is a technique used to prove results statements for natural numbers using a domino effect. We show that the first case is true and that if this is true then the next case must also be true.

De Moivre’s theorem:

If $\scriptsize z=r(\cos \theta +i\sin \theta )$ is a complex number then
$\scriptsize {{z}^{n}}={{r}^{n}}\left[ {\cos (n\theta )+i\sin (n\theta )} \right]$
$\scriptsize {{z}^{n}}={{r}^{n}}\text{cis}(n\theta )$
Where $\scriptsize n$ is a positive integer.

### Example 4.1

Evaluate $\scriptsize {{\left( {\sqrt{3}+i} \right)}^{5}}$ using de Moivre’s theorem, leaving your answer in standard form.

Solution

De Moivre’s theorem applies to complex numbers in polar form, so we first need to write our number in polar form.
\scriptsize \begin{align*}z & =\sqrt{3}+i\\\therefore \left| z \right| & =\sqrt{{{{{\left( {\sqrt{3}} \right)}}^{2}}+{{1}^{2}}}}\\ & =\sqrt{{3+1}}\\=2\end{align*}
$\scriptsize z$ is in the first quadrant.
\scriptsize \begin{align*}\sin \theta & =\displaystyle \frac{1}{2}\\\therefore \theta & ={{30}^\circ}\end{align*}
$\scriptsize z=2(\cos {{30}^\circ}+i\sin {{30}^\circ})$

Now we can use de Moivre’s theorem to evaluate $\scriptsize {{\left( {\sqrt{3}+i} \right)}^{5}}$.
\scriptsize \begin{align*}{{\left( {\sqrt{3}+i} \right)}^{5}}&={{\left[ {2(\cos {{{30}}^\circ}+i\sin {{{30}}^\circ})} \right]}^{5}}\\&={{2}^{5}}\left[ {\cos (5\times {{{30}}^\circ})+i\sin (5\times {{{30}}^\circ})} \right]\\&=32(\cos {{150}^\circ}+i\sin {{150}^\circ})\\&=32\left( {-\displaystyle \frac{{\sqrt{3}}}{2}+\displaystyle \frac{1}{2}i} \right)\\&=-16\sqrt{3}+16i\end{align*}

### Example 4.2

Evaluate $\scriptsize {{\left( {2-2i} \right)}^{8}}$ using de Moivre’s theorem, leaving your answer in standard form.

Solution

De Moivre’s theorem applies to complex numbers in polar form, so we first need to write our number in polar form.
\scriptsize \begin{align*}z &=2-2i\\\therefore \left| z \right| & =\sqrt{{{{2}^{2}}+{{{(-2)}}^{2}}}}\\ & =\sqrt{{4+4}}\\&=\sqrt{8}\\&=2\sqrt{2}\end{align*}
$\scriptsize z$ is in the fourth quadrant.

\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{2}{{2\sqrt{2}}}=\displaystyle \frac{1}{{\sqrt{2}}}\\\therefore \alpha & ={{45}^\circ}\end{align*}
$\scriptsize \theta ={{360}^\circ}-{{45}^\circ}={{315}^\circ}$
$\scriptsize z=2\sqrt{2}\text{cis}{{315}^\circ}$

Now we can use de Moivre’s theorem to evaluate $\scriptsize {{\left( {2-2i} \right)}^{8}}$.
\scriptsize \begin{align*}{{\left( {2-2i} \right)}^{8}}&={{\left( {2\sqrt{2}\text{cis}{{{315}}^\circ}} \right)}^{8}}\\&={{\left( {2\sqrt{2}} \right)}^{8}}\left[ {\text{cis}(8\times {{{315}}^\circ})} \right]\\&=256\times 16\text{cis}2\ {{520}^\circ}\\&=4\ 096\text{cis}2\ {{520}^\circ}\end{align*}

It is better to leave your answer with an angle that is between zero and $\scriptsize {{360}^\circ}$ by adding or subtracting the necessary multiple of $\scriptsize {{360}^\circ}$ from it.

Therefore, $\scriptsize {{\left( {2-2i} \right)}^{8}}=4\ 096\text{cis}{{0}^\circ}$.

### Example 4.3

Use de Moivre’s theorem to evaluate $\scriptsize \displaystyle \frac{{{{{(3+2i)}}^{6}}}}{{5-3i}}$, leaving your answer in polar form.

Solution

Remember that to use de Moivre’s theorem, the complex number must be in polar form. We need to convert $\scriptsize 3+2i$ to polar form to evaluate the numerator. In questions like this, where we need to work separately with different complex numbers in the expression, it is a good idea to designate each part of the expression as a differently numbered complex number.

In $\scriptsize \displaystyle \frac{{{{{(3+2i)}}^{6}}}}{{5-3i}}$, let $\scriptsize {{z}_{1}}=3+2i$ and $\scriptsize {{z}_{2}}=5-3i$

$\scriptsize {{z}_{1}}=3+2i$
\scriptsize \begin{align*}\left| {{{z}_{1}}} \right|&=\sqrt{{{{3}^{2}}+{{2}^{2}}}}\\&=\sqrt{{9+4}}\\&=\sqrt{{13}}\end{align*}
$\scriptsize {{z}_{1}}$ is in the first quadrant.
\scriptsize \begin{align*}\sin {{\theta }_{1}} & =\displaystyle \frac{2}{{\sqrt{{13}}}}\\\therefore {{\theta }_{1}} & ={{33.69}^\circ}\end{align*}
$\scriptsize {{z}_{1}}=\sqrt{{13}}\text{cis}{{33.69}^\circ}$

Now we can evaluate $\scriptsize {{(3+2i)}^{6}}$.
\scriptsize \begin{align*}{{({{z}_{1}})}^{6}}&={{(3+2i)}^{6}}\\&={{\left[ {\sqrt{{13}}\text{cis}{{{33.69}}^\circ}} \right]}^{6}}\\&={{\left( {\sqrt{{13}}} \right)}^{6}}\left[ {\text{cis}(6\times {{{33.69}}^\circ})} \right]\\&=2\ 197\text{cis}{{202.14}^\circ}\\&=2\ 197(\cos {{202.14}^\circ}+i\sin {{202.14}^\circ})\end{align*}

At this point, we can either convert the numerator into standard form, or convert the denominator into polar form. Dividing complex numbers in polar form tends to be easier so converting everything to polar form is normally the best option.

$\scriptsize {{z}_{2}}=5-3i$
\scriptsize \begin{align*}\left| {{{z}_{2}}} \right|&=\sqrt{{{{5}^{2}}+{{{(-3)}}^{2}}}}\\&=\sqrt{{25+9}}\\&=\sqrt{{34}}\end{align*}
$\scriptsize {{z}_{2}}$ is in the fourth quadrant.

\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{3}{{\sqrt{{34}}}}\\\therefore \alpha & ={{30.96}^\circ}\end{align*}
$\scriptsize {{\theta }_{2}}={{360}^\circ}-{{30.96}^\circ}={{329.04}^\circ}$
$\scriptsize {{z}_{2}}=\sqrt{{34}}\text{cis}{{329.04}^\circ}$

Therefore:
\scriptsize \begin{align*}\displaystyle \frac{{{{{(3+2i)}}^{6}}}}{{5-3i}}&=\displaystyle \frac{{2\ 197\text{cis}{{{202.14}}^\circ}}}{{\sqrt{{34}}\text{cis}{{{329.04}}^\circ}}}\\&=\displaystyle \frac{{2\ 197}}{{\sqrt{{34}}}}\text{cis}({{202.14}^\circ}-{{329.04}^\circ})\\&=\displaystyle \frac{{2\ 197}}{{\sqrt{{34}}}}\text{cis}(-{{126.9}^\circ})\end{align*}

You can leave you answer with a negative angle, but it is better to change it into a positive angle by adding the necessary multiple of $\scriptsize {{360}^\circ}$.
$\scriptsize \displaystyle \frac{{{{{(3+2i)}}^{6}}}}{{5-3i}}=\displaystyle \frac{{2\ 197}}{{\sqrt{{34}}}}\text{cis}{{233.1}^\circ}$

### Exercise 4.1

Use de Moivre’s theorem to evaluate the following, leaving your answer in polar form:

1. $\scriptsize {{\left( {\sqrt{3}-\sqrt{3}i} \right)}^{7}}$
2. $\scriptsize \displaystyle \frac{{{{{(4-8i)}}^{2}}}}{{{{{\left( {-3-\sqrt{3}i} \right)}}^{3}}}}$

The full solutions are at the end of the unit.

### Did you know?

De Moivre’s theorem can be expanded to deal with roots of complex numbers as well. The symmetry is truly beautiful.

If $\scriptsize z=r(\cos \theta +i\sin \theta )$ is a complex number, then:
$\scriptsize {{z}^{{\displaystyle \frac{1}{n}}}}={{r}^{{\displaystyle \frac{1}{n}}}}\left[ {\cos \left( {\displaystyle \frac{\theta }{n}} \right)+i\sin \left( {\displaystyle \frac{\theta }{n}} \right)} \right]$
$\scriptsize {{z}^{{\displaystyle \frac{1}{n}}}}={{r}^{{\displaystyle \frac{1}{n}}}}\text{cis}\left( {\displaystyle \frac{\theta }{n}} \right)$
Where $\scriptsize n$ is a positive integer.

However, fractional powers are excluded from the NC(V) curriculum.

## Summary

In this unit you have learnt the following:

• What de Moivre’s theorem is.
• How to use de Moivre’s theorem to raise complex numbers to exponents

# Unit 4: Assessment

#### Suggested time to complete: 45 minutes

Use de Moivre’s theorem to evaluate the following and leave your answer in polar form:

1. $\scriptsize {{\left( {-\sqrt{2}+\sqrt{2}i} \right)}^{6}}$
2. $\scriptsize \displaystyle \frac{{{{{(2+5i)}}^{3}}}}{{(1-2i)}}$
3. $\scriptsize \displaystyle \frac{{{{{(4-8i)}}^{3}}}}{{{{{\left( {\sqrt{3}+3i} \right)}}^{4}}}}$
4. $\scriptsize \displaystyle \frac{{(2+4i)}}{{\left( {\sqrt{2}-3i} \right){{{\left( {\sqrt{2}+3i} \right)}}^{3}}}}$

The full solutions are at the end of the unit.

# Unit 4: Solutions

### Exercise 4.1

1. $\scriptsize z=\sqrt{3}-\sqrt{3}i$
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{3}^{2}}+{{{(-3)}}^{2}}}}\\&=\sqrt{{9+9}}\\&=\sqrt{{18}}\\&=3\sqrt{2}\end{align*}
$\scriptsize z$ is in the fourth quadrant.
\scriptsize \begin{align*}\sin \alpha &=\displaystyle \frac{{\sqrt{3}}}{{3\sqrt{2}}}\\\therefore \alpha &={{24.09}^\circ}\end{align*}
$\scriptsize \theta ={{360}^\circ}-{{24.09}^\circ}={{335.91}^\circ}$
$\scriptsize z=3\sqrt{2}\text{cis}{{335.91}^\circ}$
\scriptsize \begin{align*}{{z}^{7}}&={{\left( {3\sqrt{2}\text{cis}{{{335.91}}^\circ}} \right)}^{7}}\\&={{\left( {3\sqrt{2}} \right)}^{7}}\text{cis}(7\times {{335.91}^\circ})\\&=\left( {2\ 187\times 8\sqrt{2}} \right)\text{cis}(2\ {{351.37}^\circ})\\&=17\ 496\sqrt{2}\text{cis191}\text{.3}{{\text{7}}^\circ}\end{align*}
2. $\scriptsize \displaystyle \frac{{{{{(4-8i)}}^{2}}}}{{{{{\left( {-3-\sqrt{3}i} \right)}}^{3}}}}$
Let $\scriptsize {{z}_{1}}=4-8i$ and $\scriptsize {{z}_{2}}=3-\sqrt{3}i$
$\scriptsize {{z}_{1}}=4-8i$
\scriptsize \begin{align*}\left| {{{z}_{1}}} \right|&=\sqrt{{{{4}^{2}}+{{{(-8)}}^{2}}}}\\&=\sqrt{{16+64}}\\&=\sqrt{{80}}\\&=4\sqrt{5}\end{align*}
$\scriptsize {{z}_{1}}$ is in the fourth quadrant.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{8}{{4\sqrt{5}}}\\\therefore \alpha & ={{63.43}^\circ}\end{align*}
$\scriptsize \theta ={{360}^\circ}-{{63.43}^\circ}={{296.57}^\circ}$
$\scriptsize {{z}_{1}}=4\sqrt{5}\text{cis}{{296.57}^\circ}$
\scriptsize \begin{align*}{{\left( {{{z}_{1}}} \right)}^{2}}&={{\left( {4\sqrt{5}\text{cis}{{{296.57}}^\circ}} \right)}^{2}}\\&={{\left( {4\sqrt{5}} \right)}^{2}}\text{cis}(2\times {{296.57}^\circ})\\&=\left( {16\times 5} \right)\text{cis}{{593.14}^\circ}\\&=80\text{cis}{{593.14}^\circ}\end{align*}
.
$\scriptsize {{z}_{2}}=-3-\sqrt{3}i$
\scriptsize \begin{align*}\left| {{{z}_{2}}} \right|&=\sqrt{{{{{\left( {-3} \right)}}^{2}}+{{{\left( {-\sqrt{3}} \right)}}^{2}}}}\\&=\sqrt{{9+3}}\\&=\sqrt{{12}}\\&=2\sqrt{3}\end{align*}
$\scriptsize {{z}_{2}}$ is in the third quadrant.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{{\sqrt{3}}}{{2\sqrt{3}}}=\displaystyle \frac{1}{2}\\\therefore \alpha & ={{30}^\circ}\end{align*}
$\scriptsize \theta ={{180}^\circ}+{{30}^\circ}={{210}^\circ}$
$\scriptsize {{z}_{2}}=2\sqrt{3}\text{cis}{{210}^\circ}$
\scriptsize \begin{align*}{{\left( {{{z}_{2}}} \right)}^{3}}&={{\left( {2\sqrt{3}\text{cis}{{{210}}^\circ}} \right)}^{3}}\\&={{\left( {2\sqrt{3}} \right)}^{3}}\text{cis}(3\times {{210}^\circ})\\&=\left( {8\times 3\sqrt{3}} \right)\text{cis63}{{\text{0}}^\circ}\\&=24\sqrt{3}\text{cis63}{{\text{0}}^\circ}\end{align*}
\scriptsize \begin{align*}\displaystyle \frac{{{{{(4-8i)}}^{2}}}}{{{{{\left( {-3-\sqrt{3}i} \right)}}^{3}}}}&=\displaystyle \frac{{80\text{cis}{{{593.14}}^\circ}}}{{24\sqrt{3}\text{cis63}{{\text{0}}^\circ}}}\\&=\displaystyle \frac{{10}}{{3\sqrt{3}}}\text{cis}({{593.14}^\circ}-\text{63}{{\text{0}}^\circ})\\&=\displaystyle \frac{{10\sqrt{3}}}{9}\text{cis}(-{{36.86}^\circ})\\&=\displaystyle \frac{{10\sqrt{3}}}{9}\text{cis323}\text{.1}{{\text{4}}^\circ}\end{align*}

Back to Exercise 4.1

### Unit 4: Assessment

1. $\scriptsize z=-\sqrt{2}+\sqrt{2}i$
\scriptsize \begin{align*}\left| z \right|&=\sqrt{{{{{\left( {-\sqrt{2}} \right)}}^{2}}+{{{\left( {\sqrt{2}} \right)}}^{2}}}}\\&=\sqrt{{2+2}}\\&=\sqrt{4}\\&=2\end{align*}
$\scriptsize z$ is in the second quadrant.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{{\sqrt{2}}}{2}=\displaystyle \frac{1}{{\sqrt{2}}}\\\therefore \alpha & ={{45}^\circ}\end{align*}
$\scriptsize \theta ={{180}^\circ}-{{45}^\circ}={{135}^\circ}$
$\scriptsize z=2\text{cis}{{135}^\circ}$
\scriptsize \begin{align*}{{\left( z \right)}^{6}}&={{\left( {2\text{cis13}{{\text{5}}^\circ}} \right)}^{6}}\\&={{\left( 2 \right)}^{6}}\text{cis}(6\times {{135}^\circ})\\&=\left( {64} \right)\text{cis81}{{\text{0}}^\circ}\\&=64\text{cis9}{{\text{0}}^\circ}\end{align*}
2. $\scriptsize \displaystyle \frac{{{{{(2+5i)}}^{3}}}}{{(1-2i)}}$
Let and $\scriptsize {{z}_{2}}=1-2i$
\scriptsize \begin{align*}\left| {{{z}_{1}}} \right|&=\sqrt{{{{2}^{2}}+{{5}^{2}}}}\\&=\sqrt{{4+25}}\\&=\sqrt{{29}}\end{align*}
$\scriptsize {{z}_{1}}$ is in the first quadrant.
\scriptsize \begin{align*}\sin \theta & =\displaystyle \frac{5}{{\sqrt{{29}}}}\\\therefore \theta & ={{68.20}^\circ}\end{align*}
$\scriptsize {{z}_{1}}=\sqrt{{29}}\text{cis68}\text{.2}{{\text{0}}^\circ}$
\scriptsize \begin{align*}{{\left( {{{z}_{1}}} \right)}^{3}}&={{\left( {\sqrt{{29}}\text{cis68}\text{.2}{{\text{0}}^\circ}} \right)}^{3}}\\&={{\left( {\sqrt{{29}}} \right)}^{3}}\text{cis}(3\times {{68.20}^\circ})\\&=29\sqrt{{29}}\text{cis204}\text{.}{{\text{6}}^\circ}\end{align*}
.
\scriptsize \begin{align*}\left| {{{z}_{2}}} \right|&=\sqrt{{{{1}^{2}}+{{{(-2)}}^{2}}}}\\&=\sqrt{{1+4}}\\&=\sqrt{5}\end{align*}
$\scriptsize {{z}_{2}}$ is in the fourth quadrant.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{2}{{\sqrt{5}}}\\\therefore \alpha & ={{63.43}^\circ}\end{align*}
$\scriptsize \theta ={{360}^\circ}-{{63.43}^\circ}={{296.57}^\circ}$
$\scriptsize {{z}_{2}}=\sqrt{5}\text{cis296}\text{.5}{{\text{7}}^\circ}$
.
\scriptsize \begin{align*}\displaystyle \frac{{{{{(2+5i)}}^{3}}}}{{(1-2i)}}&=\displaystyle \frac{{29\sqrt{{29}}\text{cis204}\text{.}{{\text{6}}^\circ}}}{{\sqrt{5}\text{cis296}\text{.5}{{\text{7}}^\circ}}}\\&=\displaystyle \frac{{29\sqrt{{29}}}}{{\sqrt{5}}}\text{cis}({{204.6}^\circ}-{{296.57}^\circ})\\&=\displaystyle \frac{{29\sqrt{{29}}}}{{\sqrt{5}}}\text{cis}(-{{91.97}^\circ})\\&=\displaystyle \frac{{29\sqrt{{29}}}}{{\sqrt{5}}}\text{cis268}\text{.0}{{\text{3}}^\circ}\end{align*}
3. $\scriptsize \displaystyle \frac{{{{{(4-8i)}}^{3}}}}{{{{{\left( {\sqrt{3}+3i} \right)}}^{4}}}}$
Let $\scriptsize {{z}_{1}}=4-8i$ and $\scriptsize {{z}_{2}}=\sqrt{3}+3i$
\scriptsize \begin{align*}\left| {{{z}_{1}}} \right|&=\sqrt{{{{4}^{2}}+{{{(-8)}}^{2}}}}\\&=\sqrt{{16+64}}\\&=\sqrt{{80}}\\&=4\sqrt{5}\end{align*}
$\scriptsize {{z}_{1}}$ is in the fourth quadrant.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{8}{{4\sqrt{5}}}\\\therefore \alpha & ={{63.43}^\circ}\end{align*}
$\scriptsize \theta ={{360}^\circ}-{{63.43}^\circ}={{296.57}^\circ}$
$\scriptsize {{z}_{1}}=4\sqrt{5}\text{cis}{{296.57}^\circ}$
\scriptsize \begin{align*}{{\left( {{{z}_{1}}} \right)}^{3}}&={{\left( {4\sqrt{5}\text{cis}{{{296.57}}^\circ}} \right)}^{3}}\\&={{\left( {4\sqrt{5}} \right)}^{3}}\text{cis}(3\times {{296.57}^\circ})\\&=320\sqrt{5}\text{cis889}\text{.7}{{\text{1}}^\circ}\end{align*}
.
\scriptsize \begin{align*}\left| {{{z}_{2}}} \right|&=\sqrt{{{{{\left( {\sqrt{3}} \right)}}^{2}}+{{3}^{2}}}}\\&=\sqrt{{3+9}}\\&=\sqrt{{12}}\\&=2\sqrt{3}\end{align*}
$\scriptsize {{z}_{2}}$ is in the first quadrant.
\scriptsize \begin{align*}\sin \theta & =\displaystyle \frac{3}{{2\sqrt{3}}}=\displaystyle \frac{{3\sqrt{3}}}{6}=\displaystyle \frac{{\sqrt{3}}}{2}\\\therefore \theta & ={{60}^\circ}\end{align*}
$\scriptsize {{z}_{2}}=2\sqrt{3}\text{cis6}{{\text{0}}^\circ}$
\scriptsize \begin{align*}{{\left( {{{z}_{2}}} \right)}^{4}}&={{\left( {2\sqrt{3}\text{cis6}{{\text{0}}^\circ}} \right)}^{4}}\\&={{\left( {2\sqrt{3}} \right)}^{4}}\text{cis}(4\times {{60}^\circ})\\&=144\text{cis24}{{\text{0}}^\circ}\end{align*}
\scriptsize \begin{align*}\displaystyle \frac{{{{{(4-8i)}}^{3}}}}{{{{{\left( {\sqrt{3}+3i} \right)}}^{4}}}}&=\displaystyle \frac{{320\sqrt{5}\text{cis889}\text{.7}{{\text{1}}^\circ}}}{{144\text{cis24}{{\text{0}}^\circ}}}\\&=\displaystyle \frac{{20\sqrt{5}}}{9}\text{cis}({{889.71}^\circ}-{{240}^\circ})\\&=\displaystyle \frac{{20\sqrt{5}}}{9}\text{cis649}\text{.7}{{\text{1}}^\circ}\\&=\displaystyle \frac{{20\sqrt{5}}}{9}\text{cis289}\text{.7}{{\text{1}}^\circ}\end{align*}
4. .
\scriptsize \begin{align*}\displaystyle \frac{{(2+4i)}}{{\left( {\sqrt{2}-3i} \right){{{\left( {\sqrt{2}+3i} \right)}}^{3}}}}&=\displaystyle \frac{{(2+4i)}}{{\left( {\sqrt{2}-3i} \right)\left( {\sqrt{2}+3i} \right){{{\left( {\sqrt{2}+3i} \right)}}^{2}}}}\\&=\displaystyle \frac{{(2+4i)}}{{\left( {2-9{{i}^{2}}} \right){{{\left( {\sqrt{2}+3i} \right)}}^{2}}}}\\&=\displaystyle \frac{{(2+4i)}}{{11{{{\left( {\sqrt{2}+3i} \right)}}^{2}}}}\end{align*}
Let $\scriptsize {{z}_{1}}=2+4i$ and $\scriptsize {{z}_{2}}=\sqrt{2}+3i$.
\scriptsize \begin{align*}\left| {{{z}_{1}}} \right|&=\sqrt{{{{2}^{2}}+{{4}^{2}}}}\\&=\sqrt{{4+16}}\\&=\sqrt{{20}}\\&=2\sqrt{5}\end{align*}
$\scriptsize {{z}_{1}}$ is in the first quadrant.
\scriptsize \begin{align*}\sin \theta & =\displaystyle \frac{4}{{2\sqrt{5}}}\\\therefore \theta & ={{63.43}^\circ}\end{align*}
$\scriptsize {{z}_{1}}=2\sqrt{5}\text{cis}{{63.43}^\circ}$
.
\scriptsize \begin{align*}\left| {{{z}_{2}}} \right|&=\sqrt{{{{{\left( {\sqrt{2}} \right)}}^{2}}+{{3}^{2}}}}\\&=\sqrt{{2+9}}\\&=\sqrt{{11}}\end{align*}
$\scriptsize {{z}_{2}}$ is in the first quadrant.
\scriptsize \begin{align*}\sin \theta & =\displaystyle \frac{3}{{\sqrt{{11}}}}\\\therefore \theta & ={{64.76}^\circ}\end{align*}
$\scriptsize {{z}_{2}}=\sqrt{{11}}\text{cis64}\text{.7}{{\text{6}}^\circ}$
\scriptsize \begin{align*}{{\left( {{{z}_{2}}} \right)}^{2}}&={{\left( {\sqrt{{11}}\text{cis64}\text{.7}{{\text{6}}^\circ}} \right)}^{2}}\\&={{\left( {\sqrt{{11}}} \right)}^{2}}\text{cis}(2\times {{64.76}^\circ})\\&=11\text{cis129}\text{.5}{{\text{2}}^\circ}\end{align*}
\scriptsize \begin{align*}\displaystyle \frac{{(2+4i)}}{{11{{{\left( {\sqrt{2}+3i} \right)}}^{2}}}}&=\displaystyle \frac{{2\sqrt{5}\text{cis}{{{63.43}}^\circ}}}{{11\times 11\text{cis129}\text{.5}{{\text{2}}^\circ}}}\\&=\displaystyle \frac{{2\sqrt{5}\text{cis}{{{63.43}}^\circ}}}{{144\text{cis129}\text{.5}{{\text{2}}^\circ}}}\\&=\displaystyle \frac{{\sqrt{5}}}{{72}}\text{cis}({{63.43}^\circ}-{{129.52}^\circ})\\&=\displaystyle \frac{{\sqrt{5}}}{{72}}\text{cis}(-{{66.09}^\circ})\\&=\displaystyle \frac{{\sqrt{5}}}{{72}}\text{cis}{{293.91}^\circ}\end{align*}

Back to Unit 4: Assessment