Question

Convert $\frac{(1+7i)}{(2-i{)}^2}$ into polar form.

Answer 1

Let $z=\frac{(1+7i)}{(2-i{)}^2}=\frac{(1+7i)}{(4+i^2-4i)}=\frac{(1+7i)}{(3-4i)}\times \frac{(3+4i)}{(3+4i)}$
$\Rightarrow z=\frac{(1+7i)(3+4i)}{(9+16)}=\frac{-25+25i}{25}=(-1+i)$
Let its polar form be $z=r(cos\theta +isin\theta )$.
Now, $r=|z|=\sqrt{(-1{)}^2+1^2}=\sqrt{2}$
Let $\alpha$ be the acute angle, given by
$tan\alpha =\left|\frac{Im\left(z\right)}{Re\left(z\right)}\right|=\left|\frac{1}{-1}\right|=1\Rightarrow \alpha =\frac{\pi }{4}$
Clearly, the point representing z=(-1+i) is P(-1,1), which lies in the second quadrant.
$\therefore arg\left(z\right)=\theta =(\pi -\alpha )=(\pi -\frac{\pi }{4})=\frac{3\pi }{4}$
Thus, $r=|z|=\sqrt{2}and\theta =\frac{3\pi }{4}$
Hence, The required polar form is $z=\sqrt{2}(cos\frac{3\pi }{4}+isin\frac{3\pi }{4})$