Question

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

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

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

Let z=x+iy = -1+i

$\therefore$ rcos$\theta$ =-1 And rsin$\theta$ =1 $\therefore$ $\theta$=$\frac{3\pi }{4}$ and r=$\sqrt{2}$

Thus $\frac{1+7i}{(2-i{)}^2}$=$\sqrt{2}$ $[cos\frac{3\pi }{4}+isin\frac{3\pi }{4}]$

Alter: $\left|\frac{1+7i}{(2-i{)}^2}\right|$=$\left|\frac{1+7i}{3-4i}\right|$=$\sqrt{2}$

And arg $\left(\frac{1+7i}{(2-i{)}^2}\right)$=${\tan }^{-1}7-{\tan }^{-1}\left(\frac{-4}{3}\right)$

=${\tan }^{-1}7+{\tan }^{-1}\left(\frac{4}{3}\right)$=$\frac{3\pi }{4}$

$\therefore$ $\frac{1+7i}{(2-i{)}^2}$=$\sqrt{2}$ $[cos\frac{3\pi }{4}+isin\frac{3\pi }{4}]$