Question

Express the complex no $\frac{1+7i}{(2-i{)}^2}$ in the polar form.

Answer 1

$\frac{1+7i}{(2-i{)}^2}=\frac{1+7i}{(2-i{)}^2}$x$\frac{(2+i{)}^2}{(2+i{)}^2}$
$=\frac{(1+7i)(2+i{)}^2}{(2-i{)}^2(2+i{)}^2}$

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

$=\frac{(4-1+4i+28i-7i-28)}{\left(\right(2-i\left)\right(2+i){)}^2}$
$=\frac{(-25+25i)}{(2^2+1{)}^2}$ $\left[\right(a+ib\left)\right(a-ib)=a^2+b^2]$
$=\frac{(-25+25i)}{25}$
$=-1+i$
If $\theta$ is principal argument and r is magnitude of complex number z then Polar form is represented by:
$z=r(\cos \theta +i\sin \theta )$
On comparision:
$-1=r\cos \theta$ and $1=r\sin \theta$
On squaring and adding we get:
$r^2({\cos }^2\theta +{\sin }^2\theta )=(-1{)}^2+1^2=2$
$r^2=2$ $[{\cos }^2\theta +{\sin }^2\theta =1]$
$r=\sqrt{2}$
further $\frac{r\sin \theta }{r\cos \theta }=\frac{1}{-1}$
$\tan \theta =-1=-\tan \left(\frac{\pi }{4}\right)$
$\theta =\frac{3\pi }{4}$ $\left[\tan \right(\pi -\theta )=-\tan \theta ]$
Polar representation of the given complex no. is:
$\frac{1+7i}{(2-i{)}^2}=-1+i=\sqrt{2}(\cos \left(\frac{3\pi }{4}\right)+i\sin \left(\frac{3\pi }{4}\right))$