Complex Numbers: Working with complex numbers

# Unit 3: Revise the multiplication and division of complex numbers in polar form

Dylan Busa

### Unit outcomes

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

• Multiply complex numbers in polar form.
• Divide complex numbers in polar form.

## What you should know

Before you start this unit, make sure you can:

• 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 form.

Refer to unit 2 in this subject outcome if you need help with any of the above.

## Introduction

In the previous unit we revised the polar form of complex numbers and how to convert between the standard (or rectangular) form and the polar form. One of the benefits of polar form is that it makes the multiplication and division of complex numbers quite easy.

## Multiply complex numbers in polar form

When we multiply complex numbers in polar form, we have to multiply the moduli and add the arguments.

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 $\scriptsize {{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)$

We could write the product using the polar form shorthand as $\scriptsize {{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\text{cis}({{\theta }_{1}}+{{\theta }_{2}})$.

### Example 3.1

Find the product of $\scriptsize {{z}_{1}}$ and $\scriptsize {{z}_{2}}$ if $\scriptsize {{z}_{1}}=3\text{cis5}{{0}^\circ}$ and $\scriptsize {{z}_{2}}=4\text{cis7}{{0}^\circ}$.

Solution

$\scriptsize {{z}_{1}}{{z}_{2}}={{r}_{1}}(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})\times {{r}_{2}}(\cos {{\theta }_{2}}+i\sin {{\theta }_{2}})={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)$
$\scriptsize {{z}_{1}}=3\text{cis5}{{0}^\circ}=3(\cos {{50}^\circ}+i\sin {{50}^\circ})$ and $\scriptsize {{z}_{2}}=4\text{cis7}{{0}^\circ}=4(\cos {{70}^\circ}+i\sin {{70}^\circ})$
\scriptsize \begin{align*}{{z}_{1}}\times {{z}_{2}}&=(3\times 4)\left( {\cos ({{{50}}^\circ}+{{{70}}^\circ})+i\sin ({{{50}}^\circ}+{{{70}}^\circ})} \right)\\&=12(\cos {{120}^\circ}+i\sin {{120}^\circ})\\&=12\text{cis}{{120}^\circ}\end{align*}

We can leave our answer in polar form or we can convert it into standard/rectangular form.
$\scriptsize \cos {{120}^\circ}=\cos ({{180}^\circ}-{{60}^\circ})=-\cos {{60}^\circ}=-\displaystyle \frac{1}{2}$
$\scriptsize \sin {{120}^\circ}=\sin ({{180}^\circ}-{{60}^\circ})=\sin {{60}^\circ}=\displaystyle \frac{{\sqrt{3}}}{2}$
\scriptsize \begin{align*}12\text{cis}{{120}^\circ}&=12\left( {-\displaystyle \frac{1}{2}+i\displaystyle \frac{{\sqrt{3}}}{2}} \right)\\&=-6+6\sqrt{3}i\end{align*}

### Example 3.2

Find the product $\scriptsize {{z}_{1}}{{z}_{2}}$, given $\scriptsize {{z}_{1}}=4(\cos {{75}^\circ}+i\sin {{75}^\circ})$and $\scriptsize {{z}_{2}}=2(\cos {{155}^\circ}+i\sin {{155}^\circ})$.

Solution

$\scriptsize {{z}_{1}}{{z}_{2}}={{r}_{1}}(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})\times {{r}_{2}}(\cos {{\theta }_{2}}+i\sin {{\theta }_{2}})={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)$
$\scriptsize {{z}_{1}}=4(\cos {{75}^\circ}+i\sin {{75}^\circ})$and $\scriptsize {{z}_{2}}=2(\cos {{155}^\circ}+i\sin {{155}^\circ})$
\scriptsize \begin{align*}{{z}_{1}}\times {{z}_{2}}&=(4\times 2)\left( {\cos ({{{75}}^\circ}+{{{155}}^\circ})+i\sin ({{{75}}^\circ}+{{{155}}^\circ})} \right)\\&=8(\cos {{230}^\circ}+i\sin {{230}^\circ})\\&=8\text{cis23}{{\text{0}}^\circ}\end{align*}

We can leave our answer in polar form or we can convert it into standard/rectangular form.
$\scriptsize \cos {{230}^\circ}=-0.643$
$\scriptsize \sin {{230}^\circ}=-0.766$
\scriptsize \begin{align*}8(\cos {{230}^\circ}+i\sin {{230}^\circ})&=8\left( {-0.643-0.766i} \right)\\&=-5.144-6.128i\end{align*}

Products of complex numbers in polar form:
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 $\scriptsize {{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)$
$\scriptsize {{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\text{cis}({{\theta }_{1}}+{{\theta }_{2}})$

### Exercise 3.1

1. Find $\scriptsize {{z}_{1}}{{z}_{2}}$ leaving your answer in polar form:
1. $\scriptsize {{z}_{1}}=2\sqrt{3}\text{cis}{{116}^\circ};\text{ }{{z}_{2}}=2\text{cis10}{{\text{9}}^\circ}$
2. $\scriptsize {{z}_{1}}=\sqrt{2}\text{cis}{{210}^\circ};\text{ }{{z}_{2}}=2\sqrt{2}\text{cis}{{100}^\circ}$
2. Find $\scriptsize {{z}_{1}}{{z}_{2}}$ leaving your answer in standard form:
1. $\scriptsize {{z}_{1}}=\displaystyle \frac{1}{3}\text{cis}{{120}^\circ};\text{ }{{z}_{2}}=4\text{cis}{{15}^\circ}$
2. $\scriptsize {{z}_{1}}=\sqrt{3}\text{cis}{{120}^\circ};\text{ }{{z}_{2}}=\displaystyle \frac{1}{{\sqrt{2}}}\text{cis}{{60}^\circ}$

The full solutions are at the end of the unit.

## Divide complex numbers in polar form

Dividing complex numbers in polar form is very similar to finding the product except that we find the quotient of the two moduli and the difference of the two arguments.

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
$\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)$

We could write the product using the polar form shorthand as $\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\text{cis}({{\theta }_{1}}-{{\theta }_{2}})$.

### Example 3.3

Determine $\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}$ if $\scriptsize {{z}_{1}}=\sqrt{3}\text{cis21}{{\text{3}}^\circ}$ and $\scriptsize {{z}_{2}}=2\text{cis3}{{\text{3}}^\circ}$.

Solution

$\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}=\displaystyle \frac{{{{r}_{1}}(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})}}{{{{r}_{2}}(\cos {{\theta }_{2}}+i\sin {{\theta }_{2}})}}=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)$
$\scriptsize {{z}_{1}}=\sqrt{3}\text{cis21}{{\text{3}}^\circ}=\sqrt{3}(\cos {{213}^\circ}+i\sin {{213}^\circ})$ and $\scriptsize {{z}_{2}}=2\text{cis3}{{\text{3}}^\circ}=2(\cos {{33}^\circ}+i\sin {{33}^\circ})$
\scriptsize \begin{align*}\displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\displaystyle \frac{{\sqrt{3}}}{2}\left( {\cos ({{{213}}^\circ}-{{{33}}^\circ})+i\sin ({{{213}}^\circ}-{{{33}}^\circ})} \right)\\&=\displaystyle \frac{{\sqrt{3}}}{2}(\cos {{180}^\circ}+i\sin {{180}^\circ})\\&=\displaystyle \frac{{\sqrt{3}}}{2}\text{cis}{{180}^\circ}\end{align*}

We can leave our answer in polar form or we can convert it into standard/rectangular form.
\scriptsize \begin{align*}\displaystyle \frac{{\sqrt{3}}}{2}(\cos {{180}^\circ}+i\sin {{180}^\circ})&=\displaystyle \frac{{\sqrt{3}}}{2}\left( {-1+i0} \right)\quad \cos {{180}^\circ}=-1\text{, }\sin {{180}^\circ}=0\\&=-\displaystyle \frac{{\sqrt{3}}}{2}+0i\\&=-\displaystyle \frac{{\sqrt{3}}}{2}\end{align*}

### Example 3.4

Find the quotient $\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}$ , given $\scriptsize {{z}_{1}}=2\sqrt{3}(\cos {{352}^\circ}+i\sin {{352}^\circ})$and $\scriptsize {{z}_{2}}=2(\cos {{52}^\circ}+i\sin {{52}^\circ})$.

Solution

$\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}=\displaystyle \frac{{{{r}_{1}}(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})}}{{{{r}_{2}}(\cos {{\theta }_{2}}+i\sin {{\theta }_{2}})}}=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)$
$\scriptsize {{z}_{1}}=2\sqrt{3}(\cos {{352}^\circ}+i\sin {{352}^\circ})$ and $\scriptsize {{z}_{2}}=2(\cos {{52}^\circ}+i\sin {{52}^\circ})$
\scriptsize \begin{align*}\displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\displaystyle \frac{{2\sqrt{3}}}{2}\left( {\cos ({{{352}}^\circ}-{{{52}}^\circ})+i\sin ({{{352}}^\circ}-{{{52}}^\circ})} \right)\\&=\sqrt{3}(\cos {{300}^\circ}+i\sin {{300}^\circ})\\&=\sqrt{3}\text{cis30}{{\text{0}}^\circ}\end{align*}

We can leave our answer in polar form or we can convert it into standard/rectangular form.
$\scriptsize \cos {{300}^\circ}=\cos ({{360}^\circ}-{{60}^\circ})=\cos {{60}^\circ}=\displaystyle \frac{1}{2}$
$\scriptsize \sin {{300}^\circ}=\sin ({{360}^\circ}+{{60}^\circ})=-\sin {{60}^\circ}=-\displaystyle \frac{{\sqrt{3}}}{2}$
\scriptsize \begin{align*}\sqrt{3}(\cos {{300}^\circ}+i\sin {{300}^\circ})&=\sqrt{3}\left( {\displaystyle \frac{1}{2}-\displaystyle \frac{{\sqrt{3}}}{2}i} \right)\\&=\displaystyle \frac{{\sqrt{3}}}{2}-\displaystyle \frac{3}{2}i\end{align*}

Quotients of complex numbers in polar form:

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
$\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)$
$\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\text{cis}({{\theta }_{1}}-{{\theta }_{2}})$

### Exercise 3.2

1. Find $\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}$ leaving your answer in polar form:
1. $\scriptsize {{z}_{1}}=7\text{cis}{{155}^\circ};\text{ }{{z}_{2}}=3\text{cis2}{{\text{0}}^\circ}$
2. $\scriptsize {{z}_{1}}=\sqrt{2}\text{cis}{{215}^\circ};\text{ }{{z}_{2}}=2\sqrt{2}\text{cis}{{100}^\circ}$
2. Find $\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}$ leaving your answer in standard form:
1. $\scriptsize {{z}_{1}}=\sqrt{3}\text{cis}{{90}^\circ};\text{ }{{z}_{2}}=\sqrt{3}\text{cis}{{60}^\circ}$
2. $\scriptsize {{z}_{1}}=12\text{cis27}{{\text{0}}^\circ};\text{ }{{z}_{2}}=3\sqrt{2}\text{cis}{{45}^\circ}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to multiply complex numbers in polar form.
• How to divide complex numbers in polar form.

# Unit 3: Assessment

#### Suggested time to complete: 45 minutes

1. Determine $\scriptsize 8\text{cis7}{{0}^\circ}\times 3\text{cis2}{{\text{5}}^\circ}$, leaving your answer in polar form.
2. Find the product of $\scriptsize {{z}_{1}}=4\text{cis}{{80}^\circ}$ and $\scriptsize {{z}_{2}}=2\text{cis14}{{\text{5}}^\circ}$, leaving your answer in polar form.
3. Find the quotient of $\scriptsize {{z}_{1}}=2\text{cis22}{{\text{3}}^\circ}$and $\scriptsize {{z}_{2}}=4\text{cis4}{{\text{3}}^\circ}$, leaving your answer in rectangular form.
4. Simplify the following complex expression, leaving your answer in polar form:
$\scriptsize \displaystyle \frac{{2.5\text{cis}{{{35}}^\circ}\times 4\text{cis}{{{45}}^\circ}}}{{2\text{cis}{{{60}}^\circ}}}$
5. Simplify the following complex expression without a calculator, leaving your answer in standard form:
$\scriptsize \displaystyle \frac{{5\text{cis17}{{5}^\circ}\times 3\sqrt{2}\text{cis1}{{{45}}^\circ}}}{{\sqrt{6}\text{cis}{{\text{5}}^\circ}}}$
6. An electrical circuit has impedances that are expressed as complex numbers $\scriptsize {{Z}_{A}}=3-3i$ and $\scriptsize {{Z}_{B}}=5+3i$. Calculate the value of the impedance $\scriptsize {{Z}_{{AB}}}$ if $\scriptsize {{Z}_{{AB}}}=\displaystyle \frac{{{{Z}_{A}}\times {{Z}_{B}}}}{{{{Z}_{A}}+{{Z}_{B}}}}$ and leave your answer in polar form.

The full solutions are at the end of the unit.

# Unit 3: Solutions

### Exercise 3.1

1. .
1. $\scriptsize {{z}_{1}}=2\sqrt{3}\text{cis}{{116}^\circ};\text{ }{{z}_{2}}=2\text{cis10}{{\text{9}}^\circ}$
\scriptsize \begin{align*}{{z}_{1}}{{z}_{2}}&={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)\\&=4\sqrt{3}(\cos {{225}^\circ}+i\sin {{225}^\circ})\\&=4\sqrt{3}\text{cis}{{225}^\circ}\end{align*}
2. $\scriptsize {{z}_{1}}=\sqrt{2}\text{cis}{{210}^\circ};\text{ }{{z}_{2}}=2\sqrt{2}\text{cis}{{100}^\circ}$
\scriptsize \begin{align*}{{z}_{1}}{{z}_{2}}&={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)\\&=4(\cos {{310}^\circ}+i\sin {{310}^\circ})\\&=4\text{cis31}{{\text{0}}^\circ}\end{align*}
2. .
1. $\scriptsize {{z}_{1}}=\displaystyle \frac{1}{3}\text{cis}{{120}^\circ};\text{ }{{z}_{2}}=4\text{cis}{{15}^\circ}$
\scriptsize \begin{align*}{{z}_{1}}{{z}_{2}}&={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)\\&=\displaystyle \frac{4}{3}(\cos {{135}^\circ}+i\sin {{135}^\circ})\\&=\displaystyle \frac{4}{3}\text{cis13}{{\text{5}}^\circ}\end{align*}
$\scriptsize \cos {{135}^\circ}=\cos ({{180}^\circ}-{{45}^\circ})=-\cos {{45}^\circ}=-\displaystyle \frac{1}{{\sqrt{2}}}$
$\scriptsize \sin {{135}^\circ}=\sin ({{180}^\circ}-{{45}^\circ})=\sin {{45}^\circ}=\displaystyle \frac{1}{{\sqrt{2}}}$
\scriptsize \begin{align*}z&=\displaystyle \frac{4}{3}\left( {-\displaystyle \frac{1}{{\sqrt{2}}}+\displaystyle \frac{1}{{\sqrt{2}}}i} \right)\\&=-\displaystyle \frac{4}{{3\sqrt{2}}}+\displaystyle \frac{4}{{3\sqrt{2}}}i\\&=-\displaystyle \frac{{4\sqrt{2}}}{6}+\displaystyle \frac{{4\sqrt{2}}}{6}i\\&=-\displaystyle \frac{{2\sqrt{2}}}{3}+\displaystyle \frac{{2\sqrt{2}}}{3}i\end{align*}
2. $\scriptsize {{z}_{1}}=\sqrt{3}\text{cis}{{120}^\circ};\text{ }{{z}_{2}}=\displaystyle \frac{1}{{\sqrt{2}}}\text{cis}{{60}^\circ}$
\scriptsize \begin{align*}{{z}_{1}}{{z}_{2}}&={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)\\&=\displaystyle \frac{{\sqrt{3}}}{{\sqrt{2}}}(\cos {{180}^\circ}+i\sin {{180}^\circ})\\&=\displaystyle \frac{{\sqrt{3}}}{{\sqrt{2}}}\text{cis18}{{\text{0}}^\circ}\end{align*}
$\scriptsize \cos {{180}^\circ}=-1$
$\scriptsize \sin {{180}^\circ}=0$
\scriptsize \begin{align*}z&=\displaystyle \frac{{\sqrt{3}}}{{\sqrt{2}}}\left( {-1+0i} \right)\\&=-\displaystyle \frac{{\sqrt{3}}}{{\sqrt{2}}}\quad \text{Rationalise the denominator by multiplying by }\displaystyle \frac{{\sqrt{2}}}{{\sqrt{2}}}\\&=-\displaystyle \frac{{\sqrt{2}\sqrt{3}}}{2}\\&=-\displaystyle \frac{{\sqrt{6}}}{2}\end{align*}

Back to Exercise 3.1

### Exercise 3.2

1. .
1. $\scriptsize {{z}_{1}}=7\text{cis}{{155}^\circ};\text{ }{{z}_{2}}=3\text{cis2}{{\text{0}}^\circ}$
\scriptsize \begin{align*}\displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)\\&=\displaystyle \frac{7}{3}(\cos {{135}^\circ}+i\sin {{135}^\circ})\\&=\displaystyle \frac{7}{3}\text{cis13}{{\text{5}}^\circ}\end{align*}
2. $\scriptsize {{z}_{1}}=\sqrt{2}\text{cis}{{215}^\circ};\text{ }{{z}_{2}}=2\sqrt{2}\text{cis}{{100}^\circ}$
\scriptsize \begin{align*}\displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)\\&=\displaystyle \frac{{\sqrt{2}}}{{2\sqrt{2}}}(\cos {{115}^\circ}+i\sin {{115}^\circ})\\&=\displaystyle \frac{1}{2}\text{cis11}{{\text{5}}^\circ}\end{align*}
2. .
1. $\scriptsize {{z}_{1}}=\sqrt{3}\text{cis}{{90}^\circ};\text{ }{{z}_{2}}=\sqrt{3}\text{cis}{{60}^\circ}$
\scriptsize \begin{align*}\displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)\\&=\displaystyle \frac{{\sqrt{3}}}{{\sqrt{3}}}(\cos {{30}^\circ}+i\sin {{30}^\circ})\\&=1\text{cis3}{{\text{0}}^\circ}\end{align*}
$\scriptsize \cos {{30}^\circ}=\displaystyle \frac{{\sqrt{3}}}{2}$
$\scriptsize \sin {{30}^\circ}=\displaystyle \frac{1}{2}$
\scriptsize \begin{align*}z&=1\left( {\displaystyle \frac{{\sqrt{3}}}{2}+\displaystyle \frac{1}{2}i} \right)\\&=\displaystyle \frac{{\sqrt{3}}}{2}+\displaystyle \frac{1}{2}i\end{align*}
2. $\scriptsize {{z}_{1}}=12\text{cis27}{{\text{0}}^\circ};\text{ }{{z}_{2}}=3\sqrt{2}\text{cis}{{45}^\circ}$
\scriptsize \begin{align*}\displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)\\&=\displaystyle \frac{{12}}{{3\sqrt{2}}}(\cos {{225}^\circ}+i\sin {{225}^\circ})\\&=\displaystyle \frac{4}{{\sqrt{2}}}\text{cis22}{{\text{5}}^\circ}\\&=\displaystyle \frac{{4\sqrt{2}}}{2}\text{cis}{{225}^\circ}\\&=2\sqrt{2}\text{cis}{{225}^\circ}\end{align*}
$\scriptsize \cos {{225}^\circ}=\cos ({{180}^\circ}+{{45}^\circ})=-\cos {{45}^\circ}=-\displaystyle \frac{1}{{\sqrt{2}}}$
$\scriptsize \sin {{225}^\circ}=\sin ({{180}^\circ}+{{45}^\circ})=-\sin {{45}^\circ}=-\displaystyle \frac{1}{{\sqrt{2}}}$
\scriptsize \begin{align*}z&=2\sqrt{2}\left( {-\displaystyle \frac{1}{{\sqrt{2}}}-\displaystyle \frac{1}{{\sqrt{2}}}i} \right)\\&=-\displaystyle \frac{{2\sqrt{2}}}{{\sqrt{2}}}-\displaystyle \frac{{2\sqrt{2}}}{{\sqrt{2}}}i\\&=-2-2i\end{align*}

Back to Exercise 3.2

### Unit 3: Assessment

1. $\scriptsize 8\text{cis7}{{0}^\circ}\times 3\text{cis2}{{\text{5}}^\circ}=24\text{cis}{{95}^\circ}$
2. $\scriptsize {{z}_{1}}=4\text{cis}{{80}^\circ}$ and $\scriptsize {{z}_{2}}=2\text{cis14}{{\text{5}}^\circ}$
\scriptsize \begin{align*}{{z}_{1}}{{z}_{2}}&={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)\\&=8(\cos {{225}^\circ}+i\sin {{225}^\circ})\\&=8\text{cis22}{{\text{5}}^\circ}\end{align*}
3. $\scriptsize {{z}_{1}}=2\text{cis22}{{\text{3}}^\circ}$and $\scriptsize {{z}_{2}}=4\text{cis4}{{\text{3}}^\circ}$
\scriptsize \begin{align*}\displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)\\&=\displaystyle \frac{1}{2}\left( {\cos ({{{223}}^\circ}-{{{43}}^\circ})+i\sin ({{{223}}^\circ}-{{{43}}^\circ})} \right)\\&=\displaystyle \frac{1}{2}(\cos {{180}^\circ}+i\sin {{180}^\circ})\\&=\displaystyle \frac{1}{2}\text{cis18}{{\text{0}}^\circ}\end{align*}
$\scriptsize \cos {{180}^\circ}=-1$
$\scriptsize \sin {{180}^\circ}=0$
\scriptsize \begin{align*}z&=\displaystyle \frac{1}{2}\left( {-1+0i} \right)\\&=-\displaystyle \frac{1}{2}\end{align*}
4. .
\scriptsize \begin{align*}\displaystyle \frac{{2.5\text{cis}{{{35}}^\circ}\times 4\text{cis}{{{45}}^\circ}}}{{2\text{cis}{{{60}}^\circ}}}&=\displaystyle \frac{{10\text{cis}{{{80}}^\circ}}}{{2\text{cis}{{{60}}^\circ}}}\\&=5\text{cis2}{{0}^\circ}\end{align*}
5. .
\scriptsize \begin{align*}\displaystyle \frac{{5\text{cis17}{{5}^\circ}\times 3\sqrt{2}\text{cis1}{{{45}}^\circ}}}{{\sqrt{6}\text{cis}{{\text{5}}^\circ}}}&=\displaystyle \frac{{15\sqrt{2}\text{cis32}{{\text{0}}^\circ}}}{{\sqrt{6}\text{cis}{{\text{5}}^\circ}}}\\&=\displaystyle \frac{{15\sqrt{2}}}{{\sqrt{2}\sqrt{3}}}\text{cis31}{{\text{5}}^\circ}\\&=\displaystyle \frac{{15}}{{\sqrt{3}}}\text{cis31}{{\text{5}}^\circ}\end{align*}
$\scriptsize \cos {{315}^\circ}=\cos ({{360}^\circ}-{{45}^\circ})=\cos {{45}^\circ}=\displaystyle \frac{1}{{\sqrt{2}}}$
$\scriptsize \sin {{315}^\circ}=\sin ({{360}^\circ}-{{45}^\circ})=-\sin {{45}^\circ}=-\displaystyle \frac{1}{{\sqrt{2}}}$
\scriptsize \begin{align*}z&=\displaystyle \frac{{15}}{{\sqrt{3}}}\left( {\displaystyle \frac{1}{{\sqrt{2}}}-\displaystyle \frac{1}{{\sqrt{2}}}i} \right)\\&=\displaystyle \frac{{15}}{{\sqrt{6}}}-\displaystyle \frac{{15}}{{\sqrt{6}}}i\quad \text{Rationalise denominators by multiply by }\displaystyle \frac{{\sqrt{6}}}{{\sqrt{6}}}\\&=\displaystyle \frac{{15\sqrt{6}}}{6}-\displaystyle \frac{{15\sqrt{6}}}{6}i\\&=\displaystyle \frac{{5\sqrt{6}}}{2}-\displaystyle \frac{{5\sqrt{6}}}{2}i\end{align*}
6. $\scriptsize {{Z}_{A}}=3-3i$ and $\scriptsize {{Z}_{B}}=5+3i$
\scriptsize \begin{align*}{{Z}_{{AB}}}&=\displaystyle \frac{{{{Z}_{A}}\times {{Z}_{B}}}}{{{{Z}_{A}}+{{Z}_{B}}}}\\&=\displaystyle \frac{{(3-3i)(5+3i)}}{{(3-3i)+(5+3i)}}\\&=\displaystyle \frac{{15+9i-15i-9{{i}^{2}}}}{8}\\&=\displaystyle \frac{{15-6i+9}}{8}\\&=\displaystyle \frac{{24-6i}}{8}\\&=3-\displaystyle \frac{3}{4}i\end{align*}
\scriptsize \begin{align*}\left| {{{Z}_{{AB}}}} \right|&=\sqrt{{{{3}^{2}}+{{{\left( {-\displaystyle \frac{3}{4}} \right)}}^{2}}}}\\&=\sqrt{{9+\displaystyle \frac{9}{{16}}}}\\&=\sqrt{{\displaystyle \frac{{144+9}}{{16}}}}\\&=\displaystyle \frac{{\sqrt{{153}}}}{4}\end{align*}
$\scriptsize {{Z}_{{AB}}}$ is in the fourth quadrant.
\scriptsize \begin{align*}\sin \alpha & =\displaystyle \frac{{\displaystyle \frac{3}{4}}}{{\displaystyle \frac{{\sqrt{{153}}}}{4}}}=\displaystyle \frac{3}{4}\times \displaystyle \frac{4}{{\sqrt{{153}}}}=\displaystyle \frac{3}{{\sqrt{{153}}}}\\\therefore \alpha & ={{14.04}^\circ}\end{align*}
$\scriptsize \theta ={{360}^\circ}-{{14.04}^\circ}={{345.96}^\circ}$
$\scriptsize {{Z}_{{AB}}}=\displaystyle \frac{{\sqrt{{153}}}}{4}\text{cis}{{345.96}^\circ}$

Back to Unit 3: Assessment