Question

The polar form of $\frac{1+7i}{{(2-i)}^2}$ is

• A. $\sqrt{2},\frac{3\pi }{4}$
• B. $\sqrt{2},\frac{\pi }{4}$
• C. $2,\frac{3\pi }{4}$
• D. $2,\frac{\pi }{4}$

Answer 1

The correct option is A $\sqrt{2},\frac{3\pi }{4}$

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

$=\frac{(1+7i)(3+4i)}{3^2-4^2{i}^2}=\frac{-25+25i}{25}=-1+i$

$1+i=(\sqrt{2},\pi /4)$

$\therefore -1+i=(\sqrt{2},\pi -\frac{\pi }{4})=\sqrt{2},\frac{3\pi }{4}$
Ans: A