# Convert the following in the polar form : $\;\large\frac{1+7i}{(2-i)^{2}}$

$(a)\;\sqrt{2}\;[cos(\large\frac{\pi}{3})+i \;sin(\large\frac{\pi}{3})]\qquad(b)\;2\;[cos(\large\frac{\pi}{6})+i\; sin(\large\frac{\pi}{6})]\qquad(c)\;\sqrt{2}\;[cos (\large\frac{3\pi}{4})+i \;sin(\large\frac{3\pi}{4})]\qquad(d)\;2\;[cos(\large\frac{\pi}{2})+i \;sin(\large\frac{\pi}{2})]$

Answer : $\;\sqrt{2}\;[cos (\large\frac{3\pi}{4})+i \;sin(\large\frac{3\pi}{4})]$
Explanation :
$\;\large\frac{1+7i}{(2-i)^{2}} = \large\frac{1+7i}{2^{2}+i^{2} -2i}$
$= \large\frac{1+7i}{4-1-2\times2\times i}$
$= \large\frac{1+7i}{3-4i} \times \large\frac{3+4i}{3+4i}$
$= \large\frac{3+4i+21i+28i^{2}}{3^{2}+4^{2}}$
$=\large\frac{3+25i-28}{9+16} \qquad [i^{2} =-1]$
$=\large\frac{-25+25i}{25}$
$=-1+i$
$\;-1+i\;$ Let $\;r cos \theta =-1 \;$ and $\;r sin \theta = 1$
on squaring and adding , we obtain
$(r cos \theta)^{2}+(r sin \theta)^{2} = (-1)^{2}+(1)^{2}$
$r^{2}(cos^{2} \theta + sin^{2} \theta) = 1+ 1$
$r^{2} = 2\qquad [cos^{2} \theta + sin^{2} \theta=1]$
$r =\sqrt{2}\qquad [conventionally , r > 0 ]$
Therefore , $\; \sqrt{2} cos \theta = -1 \;$ and $\; \sqrt{2} sin \theta = 1$
$cos \theta = \large\frac{-1}{\sqrt{2}}\;$ and $\;sin \theta = \large\frac{1}{\sqrt{2}}$
$\theta =(\pi- \large\frac{\pi}{4}) = \large\frac{3 \pi}{4}\qquad [\theta \;lies\; in\;the\;2^{nd} \; quadrant]$
$z=r cos \theta + i sin \theta$
$=\sqrt{2}\;[cos (\large\frac{3\pi}{4})+i \;sin(\large\frac{3\pi}{4})]$
This is the polar form