Complex numbers: Perform operations on complex numbers

# Unit 2: Multiply and divide complex numbers in polar form

Dylan Busa

### Unit outcomes

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

• Perform multiplication on complex numbers in polar form.
• Perform division on complex numbers in polar form.

## What you should know

Before you start this unit, make sure you can:

• Convert complex numbers from standard/rectangular form to polar form. Refer to unit 3 in subject outcome 1.1 if you need help with this.
• Convert complex numbers from polar form to standard/rectangular form. Refer to unit 3 in subject outcome 1.1 if you need help with this.
• Recognise and calculate the basic trigonometric ratios of the three special angles in all four quadrants. Refer to unit 3 in subject outcome 1.1 if you need help with this.

## Introduction

In the previous unit, we saw how to multiply and divide complex numbers in standard or rectangular form. One of the benefits of polar form, however, is that it makes the multiplication and division of complex numbers even easier.

### Note

In level 4 you will learn that dealing with powers and roots of complex numbers is also easier in polar form than it is in standard form.

## Multiply complex numbers in polar form

Now that we can convert complex numbers to polar form, we will learn how to multiply complex numbers in polar form. The formula involved was developed by the French mathematician Abraham De Moivre (1667–1754) (shown in Figure 1).

De Moivre’s formula requires us 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 $\scriptsize {{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)$.

Can you see that we need to multiply the moduli ($\scriptsize {{r}_{1}}\times {{r}_{2}}$) and add the arguments $\scriptsize ({{\theta }_{1}}+{{\theta }_{2}})$?

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

### Did you know?

Abraham de Moivre, on reading Isaac Newton’s Principia Mathematica, realised that this book was far deeper than any he had yet read. He became so determined to read and understand it that he tore pages out of his copy in order to read them while walking around London.

### Example 2.1

Find the product of $\scriptsize {{z}_{1}}$ and $\scriptsize {{z}_{2}}$ if $\scriptsize {{z}_{1}}=3\text{cis}{{40}^\circ}$ and $\scriptsize {{z}_{2}}=4\text{cis8}{{0}^\circ}$.

Solution

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

$\scriptsize {{z}_{1}}=3\text{cis}{{40}^\circ}=3(\cos {{40}^\circ}+i\sin {{40}^\circ})$ and $\scriptsize {{z}_{2}}=4\text{cis8}{{0}^\circ}=4(\cos {{80}^\circ}+i\sin {{80}^\circ})$

\scriptsize \begin{align*}{{z}_{1}}\times {{z}_{2}}&=(3\times 4)\left( {\cos ({{{40}}^\circ}+{{{80}}^\circ})+i\sin ({{{40}}^\circ}+{{{80}}^\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 2.2

Find the product $\scriptsize {{z}_{1}}{{z}_{2}}$, given $\scriptsize {{z}_{1}}=4(\cos {{80}^\circ}+i\sin {{80}^\circ})$and $\scriptsize {{z}_{2}}=2(\cos {{145}^\circ}+i\sin {{145}^\circ})$.

Solution

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

$\scriptsize {{z}_{1}}=4(\cos {{80}^\circ}+i\sin {{80}^\circ})$and $\scriptsize {{z}_{2}}=2(\cos {{145}^\circ}+i\sin {{145}^\circ})$

\scriptsize \begin{align*}{{z}_{1}}\times {{z}_{2}}&=(4\times 2)\left( {\cos ({{{80}}^\circ}+{{{145}}^\circ})+i\sin ({{{80}}^\circ}+{{{145}}^\circ})} \right)\\&=8(\cos {{225}^\circ}+i\sin {{225}^\circ})\\&=8\text{cis22}{{\text{5}}^\circ}\end{align*}

We can leave our answer in polar form or we can convert it into standard/rectangular form.
$\scriptsize \cos {{225}^\circ}=\cos ({{180}^\circ}+{{45}^\circ})=-\cos {{45}^\circ}=-\displaystyle \frac{1}{{\sqrt{2}}}$
$\scriptsize \sin {{225}^\circ}=\sin ({{180}^\circ}+{{45}^\circ})=-\sin {{45}^\circ}=-\displaystyle \frac{1}{{\sqrt{2}}}$
\scriptsize \begin{align*}8(\cos {{225}^\circ}+i\sin {{225}^\circ})&=8\left( {-\displaystyle \frac{1}{{\sqrt{2}}}-i\displaystyle \frac{1}{{\sqrt{2}}}} \right)\\&=-\displaystyle \frac{8}{{\sqrt{2}}}-\displaystyle \frac{8}{{\sqrt{2}}}i\quad \text{Rationalise the denominators by multiplying by }\displaystyle \frac{{\sqrt{2}}}{{\sqrt{2}}}\\&=-\displaystyle \frac{{8\sqrt{2}}}{2}-\displaystyle \frac{{8\sqrt{2}}}{2}i\\&=-4\sqrt{2}-4\sqrt{2}i\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 $\scriptsize {{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\left( {\cos ({{\theta }_{1}}+{{\theta }_{2}})+i\sin ({{\theta }_{1}}+{{\theta }_{2}})} \right)$
$\scriptsize {{z}_{1}}{{z}_{2}}={{r}_{1}}{{r}_{2}}\text{cis}({{\theta }_{1}}+{{\theta }_{2}})$

The product requires multiplying the moduli and adding the angles.

### Exercise 2.1

1. Find $\scriptsize {{z}_{1}}{{z}_{2}}$ leaving your answer in polar form:
1. $\scriptsize {{z}_{1}}=2\sqrt{3}\text{cis}{{116}^\circ};\text{ }{{z}_{2}}=2\text{cis}{{82}^\circ}$
2. $\scriptsize {{z}_{1}}=\sqrt{2}\text{cis}{{205}^\circ};\text{ }{{z}_{2}}=2\sqrt{2}\text{cis}{{118}^\circ}$
2. Find $\scriptsize {{z}_{1}}{{z}_{2}}$ leaving your answer in standard form:
1. $\scriptsize {{z}_{1}}=3\text{cis}{{120}^\circ};\text{ }{{z}_{2}}=\displaystyle \frac{1}{4}\text{cis}{{15}^\circ}$
2. $\scriptsize {{z}_{1}}=3\text{cis}{{120}^\circ};\text{ }{{z}_{2}}=5\sqrt{2}\text{cis}{{60}^\circ}$

The full solutions are at the end of the unit.

## Divide complex numbers in polar form

The quotient of two 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. This is another formula developed by Abraham De Moivre.

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 $\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)$. Can you see that we need to divide the moduli $\scriptsize \left(\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\right)$ and subtract the arguments $\scriptsize ({{\theta }_{1}}-{{\theta }_{2}})$?

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

### Example 2.3

Find the quotient of $\scriptsize {{z}_{1}}$ and $\scriptsize {{z}_{2}}$ if $\scriptsize {{z}_{1}}=2\text{cis21}{{\text{3}}^\circ}$ and $\scriptsize {{z}_{2}}=4\text{cis3}{{\text{3}}^\circ}$.

Solution

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

$\scriptsize {{z}_{1}}=2\text{cis21}{{\text{3}}^\circ}=2(\cos {{213}^\circ}+i\sin {{213}^\circ})$ and $\scriptsize {{z}_{2}}=4\text{cis3}{{\text{3}}^\circ}=4(\cos {{33}^\circ}+i\sin {{33}^\circ})$

\scriptsize \begin{align*}\displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\displaystyle \frac{2}{4}\left( {\cos ({{{213}}^\circ}-{{{33}}^\circ})+i\sin ({{{213}}^\circ}-{{{33}}^\circ})} \right)\\&=\displaystyle \frac{1}{2}(\cos {{180}^\circ}+i\sin {{180}^\circ})\\&=\displaystyle \frac{1}{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{1}{2}(\cos {{180}^\circ}+i\sin {{180}^\circ})&=\displaystyle \frac{1}{2}\left( {-1+i0} \right)\quad \text{ }\cos {{180}^\circ}=-1\text{, }\sin {{180}^\circ}=0\\&=-\displaystyle \frac{1}{2}+0i\end{align*}

### Example 2.4

Find the quotient $\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}$ , given $\scriptsize {{z}_{1}}=2\sqrt{3}(\cos {{242}^\circ}+i\sin {{242}^\circ})$and $\scriptsize {{z}_{2}}=2(\cos {{32}^\circ}+i\sin {{32}^\circ})$.

Solution

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

$\scriptsize {{z}_{1}}=2\sqrt{3}(\cos {{242}^\circ}+i\sin {{242}^\circ})$ and $\scriptsize {{z}_{2}}=2(\cos {{32}^\circ}+i\sin {{32}^\circ})$

\scriptsize \begin{align*}\displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}&=\displaystyle \frac{{2\sqrt{3}}}{2}\left( {\cos ({{{242}}^\circ}-{{{32}}^\circ})+i\sin ({{{242}}^\circ}-{{{32}}^\circ})} \right)\\&=\sqrt{3}(\cos {{210}^\circ}+i\sin {{210}^\circ})\\&=\sqrt{3}\text{cis21}{{\text{0}}^\circ}\end{align*}

We can leave our answer in polar form or we can convert it into standard/rectangular form.
$\scriptsize \cos {{210}^\circ}=\cos ({{180}^\circ}+{{30}^\circ})=-\cos {{30}^\circ}=-\displaystyle \frac{{\sqrt{3}}}{2}$
$\scriptsize \sin {{210}^\circ}=\sin ({{180}^\circ}+{{30}^\circ})=-\sin {{30}^\circ}=-\displaystyle \frac{1}{2}$
\scriptsize \begin{align*}\sqrt{3}(\cos {{210}^\circ}+i\sin {{210}^\circ})&=\sqrt{3}\left( {-\displaystyle \frac{{\sqrt{3}}}{2}-\displaystyle \frac{1}{2}i} \right)\\&=-\displaystyle \frac{3}{2}-\displaystyle \frac{{\sqrt{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
$\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\left( {\cos ({{\theta }_{1}}-{{\theta }_{2}})+i\sin ({{\theta }_{1}}-{{\theta }_{2}})} \right)$
$\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}=\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}\text{cis}({{\theta }_{1}}-{{\theta }_{2}})$

The quotient requires dividing the moduli and subtracting the angles.

### Exercise 2.2

1. Find $\scriptsize \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}}$ leaving your answer in polar form:
1. $\scriptsize {{z}_{1}}=21\text{cis}{{135}^\circ};\text{ }{{z}_{2}}=3\text{cis}{{65}^\circ}$
2. $\scriptsize {{z}_{1}}=\sqrt{2}\text{cis}{{205}^\circ};\text{ }{{z}_{2}}=2\sqrt{2}\text{cis}{{118}^\circ}$
2. Find $\scriptsize {{z}_{1}}{{z}_{2}}$ leaving your answer in standard form:
1. $\scriptsize {{z}_{1}}=\sqrt{2}\text{cis}{{90}^\circ};\text{ }{{z}_{2}}=3\text{cis}{{60}^\circ}$
2. $\scriptsize {{z}_{1}}=15\text{cis}{{120}^\circ};\text{ }{{z}_{2}}=3\sqrt{2}\text{cis}{{40}^\circ}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

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

# Unit 2: Assessment

#### Suggested time to complete: 45 minutes

1. Simplify the following complex expressions, leaving your answer in polar form:
1. $\scriptsize \left( {4\text{cis}{{{60}}^\circ}} \right)\left( {6\text{cis}{{{40}}^\circ}} \right)\left( {\displaystyle \frac{1}{2}\text{cis}(-{{{20}}^\circ})} \right)$
2. $\scriptsize \displaystyle \frac{{1.25\text{cis}{{{35}}^\circ}\times 3.1\text{cis}{{{45}}^\circ}}}{{2\text{cis}{{{60}}^\circ}}}$
2. Simplify the following complex expressions, leaving your answer in standard form:
1. $\scriptsize 6(\cos {{50}^\circ}+i\sin {{50}^\circ})\times 3(\cos {{10}^\circ}+i\sin {{10}^\circ})$
2. $\scriptsize \left( {6\text{cis}{{{60}}^\circ}} \right)\left( {4\text{cis}{{{40}}^\circ}} \right)\left( {2\text{cis}{{{20}}^\circ}} \right)$
3. $\scriptsize \displaystyle \frac{{5\text{cis17}{{5}^\circ}\times 3\sqrt{2}\text{cis1}{{{45}}^\circ}}}{{\sqrt{6}\text{cis}{{\text{5}}^\circ}}}$

The full solutions are at the end of the unit.

# Unit 2: Solutions

### Exercise 2.1

1. .
1. .
\scriptsize \begin{align*} z_1&=2\sqrt{3}\text{cis}116^\circ;z_2=2\text{cis}82^\circ\\ \therefore z_1z_2&=4\sqrt{3}\text{cis}198^\circ \end{align*}
2. .
\scriptsize \begin{align*} z_1&=\sqrt{2}\text{cis}205^\circ; z_2=2\sqrt{2}\text{cis}118^\circ\\ \therefore z_1z_2&=4\text{cis}323^\circ \end{align*}
2. .
1. $\scriptsize {{z}_{1}}=3\text{cis}{{120}^\circ};\text{ }{{z}_{2}}=\displaystyle \frac{1}{4}\text{cis}{{15}^\circ}$
\scriptsize \begin{align*}\therefore {{z}_{1}}{{z}_{2}}&=\displaystyle \frac{3}{4}\text{cis}{{135}^\circ}\\ &=\displaystyle \frac{3}{4}\left( {\cos {{{135}}^\circ}+i\sin {{{135}}^\circ}} \right)&& ({{180}^\circ}-{{45}^\circ}={{135}^\circ})\\ &=\displaystyle \frac{3}{4}\left( {-\displaystyle \frac{1}{{\sqrt{2}}}+i\displaystyle \frac{1}{{\sqrt{2}}}} \right)&&\cos {{135}^\circ} \lt 0\text{ (second quadrant), sin13}{{\text{5}}^\circ} \gt 0\text{ (second quadrant)}\\ &=-\displaystyle \frac{3}{{4\sqrt{2}}}+\displaystyle \frac{3}{{4\sqrt{2}}}i&& \text{Rationalise the deonominators by multiplying by }\displaystyle \frac{{\sqrt{2}}}{{\sqrt{2}}}\\ &=-\displaystyle \frac{{3\sqrt{2}}}{8}+\displaystyle \frac{{3\sqrt{2}}}{8}i\end{align*}
2. $\scriptsize {{z}_{1}}=3\text{cis}({{120}^\circ});\text{ }{{z}_{2}}=5\sqrt{2}\text{cis}({{60}^\circ})$
\scriptsize \begin{align*}\therefore {{z}_{1}}{{z}_{2}} & =15\sqrt{2}\text{cis}{{180}^\circ}\\&=15\sqrt{2}\left( {\cos {{{180}}^\circ}+i\sin {{{180}}^\circ}} \right)\\&=15\sqrt{2}\left( {-1+0i} \right)\quad \text{ }\cos {{180}^\circ}=-1\text{, sin18}{{\text{0}}^\circ}=0\\&=-15\sqrt{2}+0i\end{align*}

Back to Exercise 2.1

### Exercise 2.2

1. .
1. .
\scriptsize \begin{align*} z_1&=21\text{cis}135^\circ; z_2=3\text{cis}65^\circ\\ \therefore \displaystyle \frac{z_1}{z_2}&=7\text{cis}70^\circ \end{align*}
2. .
\scriptsize \begin{align*} {z}_{1}&=\sqrt{2}\text{cis}205^\circ;{{z}_{2}}=2\sqrt{2}\text{cis}{{118}^\circ}\\ \therefore \displaystyle \frac{z_1}{z_2}&=\displaystyle \frac{\sqrt{2}}{2\sqrt{2}}\text{cis}87^\circ\\ &=\displaystyle \frac{1}{2}87^\circ \end{align*}
2. .
1. $\scriptsize {{z}_{1}}=\sqrt{2}\text{cis}{{90}^\circ};\text{ }{{z}_{2}}=3\text{cis}{{60}^\circ}$
\scriptsize \begin{align*}\therefore \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}} & =\displaystyle \frac{{\sqrt{2}}}{3}\text{cis3}{{\text{0}}^\circ}\\&=\displaystyle \frac{{\sqrt{2}}}{3}\left( {\cos {{{30}}^\circ}+i\sin {{{30}}^\circ}} \right)\\&=\displaystyle \frac{{\sqrt{2}}}{3}\left( {\displaystyle \frac{{\sqrt{3}}}{2}+i\displaystyle \frac{1}{2}} \right)\\&=\displaystyle \frac{{\sqrt{6}}}{6}+\displaystyle \frac{{\sqrt{2}}}{6}i\end{align*}
2. $\scriptsize {{z}_{1}}=15\text{cis}{{120}^\circ};\text{ }{{z}_{2}}=3\sqrt{2}\text{cis}{{40}^\circ}$
\scriptsize \begin{align*}\therefore \displaystyle \frac{{{{z}_{1}}}}{{{{z}_{2}}}} & =\displaystyle \frac{{15}}{{3\sqrt{2}}}\text{cis(12}{{\text{0}}^\circ}-{{40}^\circ})\\&=\displaystyle \frac{{15}}{{3\sqrt{2}}}\text{cis8}{{\text{0}}^\circ}\\&=\displaystyle \frac{5}{{\sqrt{2}}}\left( {\cos {{{80}}^\circ}+i\sin {{{80}}^\circ}} \right)\\&=\displaystyle \frac{5}{{\sqrt{2}}}\left( {0.1736+0.9848i} \right)\\&=0.6139+4.818i\end{align*}

Back to Exercise 2.2

### Unit 2: Assessment

1. .
1. .
\scriptsize \begin{align*}&\left( {4\text{cis}{{{60}}^\circ}} \right)\left( {6\text{cis}{{{40}}^\circ}} \right)\left( {\displaystyle \frac{1}{2}\text{cis}(-{{{20}}^\circ})} \right)&=\left( {4\times 6\times \displaystyle \frac{1}{2}} \right)\text{cis}\left( {{{{60}}^\circ}+{{{40}}^\circ}+(-{{{20}}^\circ})} \right)\\&=12\text{cis}{{80}^\circ}\end{align*}
2. .
\scriptsize \begin{align*} \displaystyle \frac{1.25\text{cis}35^\circ \times 3.1\text{cis}45^\circ}{2\text{cis}60^\circ}&=\displaystyle \frac{(1.25 \times 3.1\text{cis}80^\circ)}{2\text{cis}60^\circ}\\ &=\displaystyle \frac{3.875\text{cis}80^\circ}{2\text{cis}60^\circ}\\ &=\left( \displaystyle \frac{3.875}{2} \right)\text{cis}20^\circ\\ &=1.9375\text{cis}20^\circ \end{align*}
2. .
1. .
\scriptsize \begin{align*} 6(\cos50^\circ+i\sin50^\circ)\times 3(\cos10^\circ+i\sin10^\circ)&=18(\cos60^\circ+i\sin60^\circ)\\ &=18 \left(\displaystyle \frac{1}{2}+i\displaystyle \frac{\sqrt{3}}{2}\right)\\ &=9+9\sqrt{3}i \end{align*}
2. .
\scriptsize \begin{align*} (6\text{cis}60^\circ)( {4\text{cis}{{{40}}^\circ}})( {2\text{cis}{{{20}}^\circ}})&=(6\times 4\times 2)\text{cis}({{60}^\circ}+{{40}^\circ}+{{20}^\circ})\\ &=48\text{cis}{{120}^\circ}&& ({{120}^\circ}={{180}^\circ}-{{60}^\circ})\\ &=48\left( {\cos {{{120}}^\circ}+i\sin {{{120}}^\circ}} \right)\\ &=48\left( {-\displaystyle \frac{1}{2}+i\displaystyle \frac{{\sqrt{3}}}{2}} \right)&& \cos {{120}^\circ} \lt 0\text{ (second quadrant), }\sin {{120}^\circ} \gt 0\text{ (second quadrant)}\\ &=-24+24\sqrt{3}i\end{align*}
3. .
\scriptsize \begin{align*} \displaystyle \frac{5\text{cis}175\circ \times 3\sqrt{2}\text{cis}145^\circ}{\sqrt{6}\text{cis}5^\circ}&=\displaystyle \frac{15\sqrt{2}\text{cis}320^\circ}{\sqrt{6}\text{cis}5^\circ}\\ &=\displaystyle \frac{15\sqrt{2}}{\sqrt{2\times 3}}\text{cis}315^\circ &&(315^\circ=360^\circ-45^\circ)\\ &=\displaystyle \frac{15}{\sqrt{3}}\left (\cos315^\circ+isin315^\circ \right)\\ &=\displaystyle \frac{15}{\sqrt{3}}\left (\displaystyle \frac{1}{\sqrt{2}}-i\displaystyle \frac{1}{\sqrt{2}} \right)&& \cos315^\circ \gt0\text{ (fourth quadrant)},\sin315^\circ \lt0 \text{ (fourth quadrant)}\\ &=\displaystyle \frac{15}{\sqrt{6}}-i\displaystyle \frac{15}{\sqrt{6}}&&\text{Rationalise the denominators by multiplying each 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*}

Back to Unit 2: Assessment