One part of the quotient has rather high exponents and this indicates that we ought to use polar form for the calculation.

Some parts of the quotient have rather high exponents and this indicates that we ought to use polar form for the calculation.

First, we write

First, we write <math>1+i\sqrt{3}</math>, <math>1-i</math> and <math>\sqrt{3}-i</math> in polar form.

<math>1+i\sqrt{3}</math>,

<math>\text{1}-i</math>

and

<math>\sqrt{3}-i</math>

in polar form.

 

<center>[[Image:3_3_1_e.gif]] [[Image:3_3_1_e_text.gif]]</center>

[[Image:3_3_1_e.gif]] [[Image:3_3_1_e_text.gif]]

This shows that

This shows that

 

{{Displayed math||<math>\begin{align}

<math>\begin{align}

1+i\sqrt{3} &= 2\Bigl(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\Bigr)\,,\\[5pt]

& 1+i\sqrt{3}=2\left( \cos \frac{\pi }{3}+i\sin \frac{\pi }{3} \right) \\

1-i &= \sqrt{2}\Bigl(\cos\Bigl(-\frac{\pi}{4}\Bigr) + i\sin\Bigl(-\frac{\pi}{4}\Bigr)\Bigr)\,,\\[5pt]

& \text{1}-i=\sqrt{2}\left( \cos \left( -\frac{\pi }{4} \right)+i\sin \left( -\frac{\pi }{4} \right) \right) \\

\sqrt{3}-i &= 2\Bigl(\cos\Bigl(-\frac{\pi}{6}\Bigr) + i\sin\Bigl(-\frac{\pi}{6}\Bigr)\Bigr)\,\textrm{.}

& \sqrt{3}-i=2\left( \cos \left( -\frac{\pi }{6} \right)+i\sin \left( -\frac{\pi }{6} \right) \right) \\

\end{align}</math>}}

\end{align}</math>

 

Now, with the help of de Moivre's formula,

Now, with the help of de Moivre's formula,