Convert the complex number in the polar form: -1 - i

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Convert the complex number in the polar form: $ -1 - i $

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Answer

Here $ z = - 1 - i = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\;\cos \theta = - 1$ and $r\sin \theta = - 1$

Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 1 + 1$$ \Rightarrow {r^2} = 2 \Rightarrow r = \sqrt 2 $
$\therefore \sqrt 2 \cos \theta = - 1$ and $\sqrt 2 \sin \theta = - 1$
$ \Rightarrow \cos \theta = \frac{{ - 1}}{{\sqrt 2 }}$ and ${\mathrm{sinθ}=\frac{-1}{\sqrt2}}$
Since $\sin \theta $ and $\cos \theta $ are both negative
$\therefore \theta $ lies in third quadrant
$\therefore \theta = \left( { - \pi + \frac{\pi }{4}} \right) = \frac{{ - 3\pi }}{4}$
Hence polar form of z is $\sqrt 2 \left[ {\cos \left( {\frac{{ - 3\pi }}{4}} \right) + i\sin \left( {\frac{{ - 3\pi }}{4}} \right)} \right]$

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