Question

Convert the following in the polar form:

$\left(i\right)\frac{1+7i}{(2-i{)}^2}$

$\left(ii\right)\frac{1+3i}{1-2i}$

Answer 1

$\left(i\right)\frac{1+7i}{(2-i{)}^2}=\frac{1+7i}{4+i^2-4i}$

$=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}=\frac{3+4i+21i+28i^2}{9-16i^2}$

$=\frac{-25+25i}{25}=-1+i$

Let $z=-1+i=r(cos\theta +isin\theta )$

$\Rightarrow rcos\theta =-1andrsin\theta =1\dots \left(i\right)$

Squaring both sides of (i) and adding

$r^2(co{s}^2\theta +si{n}^2\theta )=1+1$

$\Rightarrow r^2=2\Rightarrow r=\sqrt{2}$

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

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

Since $sin\theta$ is positive and $cos\theta$ is negative

$\therefore \theta$ lies in the second quadrant

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

Hence polar form of z is $\sqrt{2}(cos\frac{3\pi }{4}+isin\frac{3\pi }{4})$

$\left(ii\right)\frac{1+3i}{1-2i}\times \frac{1+2i}{1+2i}=\frac{1+2i+3i+6i^2}{1-4i^2}=\frac{-5+5i}{5}=-1+i$

Let $z=-1+i=r(cos\theta +isin\theta )$

$\Rightarrow rcos\theta =-1andrsin\theta =1\dots \left(i\right)$

Squaring both sides of (i) and adding

$r^2(co{s}^2\theta +si{n}^2\theta )=1+1$

$\Rightarrow r^2=2\Rightarrow r=\sqrt{2}$

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

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

Since $sin\theta$ is positive and $cos\theta$ is negative

$\therefore \theta$ lies in second quadrant

$\therefore \theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$
Hence polar form of z is $\sqrt{2}(cos\frac{3\pi }{4}+isin\frac{3\pi }{4})$