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

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 $\sqrt 2 \sin \theta = 1$
$ {\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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