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Question
Write 1 − i in polar form.
Solution
\[z = 1 - i \]
\[r = \left| z \right|\]
\[ = \sqrt{1 + 1}\]
\[ = \sqrt{2}\]
\[\text { Let } \tan \alpha = \left| \frac{Im\left( z \right)}{Re\left( z \right)} \right|\]
\[ \therefore \tan\alpha = \left| \frac{- 1}{1} \right|\]
\[ = \frac{\pi}{4}\]
\[ \Rightarrow \alpha = \frac{\pi}{4}\]
\[\text { Since point (1, - 1) lies in the fourth quadrant, the argument of z is given by } \]
\[\theta = - \alpha = - \frac{\pi}{4}\]
\[\text { Polar form = } r\left( \cos \theta + i\sin \theta \right) \]
\[ = \sqrt{2}\left\{ \cos\left( - \frac{\pi}{4} \right) + i\sin\left( - \frac{\pi}{4} \right) \right\}\]
\[ = \sqrt{2}\left( \cos\frac{\pi}{4} - i\sin\frac{\pi}{4} \right)\]
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