Convert the following in the polar form: (i)

Q : 5 Convert the following in the polar form:

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

Answers (1)

Let
$z =\small \frac{1+7i}{(2-i)^2} = \frac{1+7i}{4+i^2-4i}= \frac{1+7i}{4-1-4i}= \frac{1+7i}{3-4i}$

Now, multiply the numerator and denominator by $3+4i$
$\Rightarrow z = \frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}= \frac{3+4i+21i+28i^2}{3^2+4^2}= \frac{-25+25i}{25}= -1+i$
Now,
let
$r\cos\theta = -1 \ \ and \ \ r\sin \theta = 1$
On squaring both and then add
$r^2(\cos^2\theta+\sin^2\theta)= (-1)^2+1^2$
$r^2=2$
$r = \sqrt2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$
Now,
$\sqrt2 \cos \theta = -1 \ \ and \ \ \sqrt2\sin \theta = 1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ and \ \ \sin \theta = \frac{1}{\sqrt2}$
Since the value of $\cos \theta$ is negative and   $\sin \theta$ is positive this is the case in II quadrant
Therefore,
$\theta = \pi - \frac{\pi}{4}= \frac{3\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
$z = r\cos \theta + ir\sin \theta$
   $=\sqrt2\cos \frac{3\pi}{4} + i\sqrt2\sin \frac{3\pi}{4}$
   $=\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )$
Therefore, the required polar form is