Question

(a) $\frac{1+7i}{{(2-i)}^2}$
(b) $2.5(\cos {300}^o+i\sin {30}^o)$

Answer 1

(a) $\frac{1+7i}{{(2-i)}^2}=\frac{1+7i}{4+i^2-4i}=\frac{1+7i}{4-1-4i}=\frac{1+7i}{3-4i}$
$\frac{(1+7i)(3+4i)}{3^2-4^2{i}^2}=\frac{-25+25i}{25}=-1+i$
$1+i=(\sqrt{2},\pi ╱4)$
$\therefore -1+i=(\sqrt{2},\pi -\frac{\pi }{4})=\sqrt{2},\frac{3\pi }{4}$
(b) First we note that $2.5\left(\cos 300+i\sin 30\right)$
$=\left(5╱2\right)[\cos (360-60)+i\sin 30]$
$=\left(5╱2\right)[\cos \left(60\right)+i\sin 30]$
$\frac{5}{2}(\frac{1}{2}+\frac{1}{2}i)=\frac{5}{4}(1+i)$
$\frac{5}{4}\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)=\frac{5}{2\sqrt{2}},\frac{\pi }{4}$