Question

Convert the following in the polar form:
(i) $\frac{1+7i}{{(2-i)}^2}$ (ii) $\frac{1+3i}{1-2i}$

Answer 1

(i) Here, $z=\frac{1+7i}{{(2-i)}^2}$
$=\frac{1+7i}{4+i^2-4i}$
$=\frac{1+7i}{4-1-4i}$
$=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}$
$=\frac{3+4i+21i+28i^2}{3^2+4^2}$
$=\frac{3+4i+21i-28}{3^2+4^2}=\frac{-25+25i}{25}$

$=-1+i$
Let $r\cos \theta =-1$ and $r\sin \theta =1$

On squaring and adding, we obtain

$r^2({\cos }^2\theta +{\sin }^2\theta )=1+1$

$\Rightarrow r^2({\cos }^2\theta +{\sin }^2\theta )=2$

$\Rightarrow r^2=2$ ($\because {\cos }^2\theta +{\sin }^2\theta =1)$

$\Rightarrow r=\sqrt{2}$ (As r>0 )

$\therefore \sqrt{2}\cos \theta =-1$ and $\sqrt{2}\sin \theta =1$

$\Rightarrow \cos \theta =\frac{-1}{\sqrt{2}}$ and

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

$\therefore \theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

(As $\theta$ lies in II quadrant )

$\therefore z=r\cos \theta +ir\sin \theta$

$=\sqrt{2}\cos \frac{3\pi }{4}+i\sqrt{2}\sin \frac{3\pi }{4}=\sqrt{2}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$

This is the required polar form.
(ii) Here,
$z=\frac{1+3i}{1-2i}$
$=\frac{1+3i}{1-2i}\times \frac{1+2i}{1+2i}$
$=\frac{1+2i+3i-6}{1+4}$
$=\frac{-5+5i}{5}$

$=-1+i$

Let $r\cos \theta =-1$ and $r\sin \theta =1$

On squaring and adding, we obtain

$r^2({\cos }^2\theta +{\sin }^2\theta )=1+1$

$\Rightarrow r^2({\cos }^2\theta +{\sin }^2\theta )=2$

$\Rightarrow r^2=2$ ($\because {\cos }^2\theta +{\sin }^2\theta =1)$

$\Rightarrow r=\sqrt{2}$ (As r>0 )

$\therefore \sqrt{2}\cos \theta =-1$ and $\sqrt{2}\sin \theta =1$

$\Rightarrow \cos \theta =\frac{-1}{\sqrt{2}}$ and

$\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}$

$\therefore \theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

(As $\theta$ lies in II quadrant )

$\therefore z=r\cos \theta +ir\sin \theta$

$=\sqrt{2}\cos \frac{3\pi }{4}+i\sqrt{2}\sin \frac{3\pi }{4}=\sqrt{2}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$

This is the required polar form.