Convert in the polar form: 1 + 3i 1 - 2i

Question

Convert in the polar form: $\frac{{1 + 3i}}{{1 - 2i}}$

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Answer

$\frac{{1 + 3i}}{{1 - 2i}} \times \frac{{1 + 2i}}{{1 + 2i}} = \frac{{1 + 2i + 3i + 6{i^2}}}{{1 - 4{i^2}}}$$ = \frac{{ - 5 + 5i}}{5} = - 1 + i$
Let $z = - 1 + i = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\cos \theta = - 1$ and $r\sin \theta = 1$
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 1 + 1$$ \Rightarrow {r^2} = 2 \Rightarrow r = \sqrt 2 $
$\therefore \sqrt 2 \cos \theta = - 1$ and $\sqrt 2 \sin \theta = 1$
$ \style{font-family:Tahoma}{\style{font-size:8px}{\Rightarrow\cos\theta=\frac{-1}{\sqrt2}}}$ and $\sin \theta = \frac{1}{{\sqrt 2 }}$
Since $\sin \theta $ is positive and $\cos \theta $ is negative
$\therefore \theta $ lies in second quadrant.
$\therefore \theta = \left( {\pi - \frac{\pi }{4}} \right) = \frac{{3\pi }}{4}$
Hence polar form of z is $\sqrt 2 \left( {\cos \frac{3\pi }{4} + i\sin \frac{{3\pi }}{4}} \right)$

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