Convert in the polar form: 1 + 7i (2 - i) ^2

Question

Convert in the polar form: $\frac{{1 + 7i}}{{{{(2 - i)}^2}}}$

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Answer

$\frac{{1 + 7i}}{{{{(2 - i)}^2}}}$$ = \frac{{1 + 7i}}{{4 + {i^2} - 4i}} = \frac{{1 + 7i}}{{3 - 4i}} \times \frac{{3 + 4i}}{{3 + 4i}}$
$ = \frac{{3 + 4i + 21i + 28{i^2}}}{{9 - 16{i^2}}}$
$ = \frac{{ - 25 + 25i}}{{25}} = - 1 + i$
Let $z = - 1 + i = r(\cos \theta + i\sin \theta )$
$ \Rightarrow r\;\cos \theta = - 1$ and $r\sin \theta = 1$ .... (i)
Squaring both sides of (i) and adding
${r^2}({\cos ^2}\theta + {\sin ^2}\theta ) = 1 + 1$$ \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 $\sin \theta = \frac{1}{{\sqrt 2 }}$
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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