Question

Polar form of $z=\frac{(1+7i)}{(2-i{)}^2}$ is

• A. $\sqrt{2}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$
• B. $2(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$
• C. $\sqrt{2}(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})$
• D. $2(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})$

Answer 1

The correct option is A $\sqrt{2}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$

Let $z=\frac{(1+7i)}{(2-i{)}^2}$
$z=\frac{1+7i}{4-4i+i^2}\Rightarrow z=\frac{1+7i}{3-4i}\Rightarrow z=\left(\frac{1+7i}{3-4i}\right)\left(\frac{3+4i}{3+4i}\right)\Rightarrow z=\frac{-25+25i}{25}\Rightarrow z=-1+i\therefore r=|z|=\sqrt{(-1{)}^2+(1{)}^2}=\sqrt{2}$
Let $\alpha$ be the acute angle given by
$\tan \alpha =\left|\frac{Im\left(z\right)}{Re\left(z\right)}\right|=|-\frac{1}{1}|=1\Rightarrow \alpha =\frac{\pi }{4}$
Since the point $(-1,1)$ representing $z$ lies in the second quadrant,
$\therefore \theta =\text{arg}\left(z\right)=\pi -\alpha =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$
Hence, $z$ in the polar form is given by
$z=\sqrt{2}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$