Question

$\frac{(1+7i)}{(2-i{)}^2}$ is equal to

• A. $\sqrt{2}[\cos (3\pi /4)+i\sin (3\pi /4\left)\right]$
• B. $\sqrt{2}[\cos (\pi /4)+i\sin (\pi /4\left)\right]$
• C. $\cos (3\pi /4)+i\sin (3\pi /4)$
• D. $Noneofthese$

Answer 1

The correct option is A

$\sqrt{2}[\cos (3\pi /4)+i\sin (3\pi /4\left)\right]$

Explanation for the correct option:

Step1. Find the value of given expression:

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

$=\frac{1+7i}{3-4i}$

Step 2. By multiplying with conjugate we get

$=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}$

$=\frac{3+4i+21i+28i^2}{3^2+4^2}$

$=\frac{3+4i+21i-28}{3^2+4^2}$

$=\frac{-25+25i}{25}$

$=-1+i$
Let $r\cos \theta =-1$ and $r\sin \theta =1$

Step 3. On squaring and adding, we get

$r^2({\cos }^2\theta +{\sin }^2\theta )=1+1$

$\Rightarrow r^2({\cos }^2\theta +{\sin }^2\theta )=2$

$\Rightarrow$ $r^2=2$ $\left[\mathbf{\because }\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{+}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{=}\mathbf{1}\right]$

$\Rightarrow$ $r=\sqrt{2}$

$\therefore$$\sqrt{2}\cos \theta =-1$and $\sqrt{2}\sin \theta =1$

$\Rightarrow$ $\cos \theta =\frac{-1}{\sqrt{2}}$​ and $\sin \theta =\frac{1}{2}$

$\Rightarrow$ $\theta =\pi -\frac{\pi }{4}$

$=\frac{3\pi }{4}$

(As θ lies in II quadrant )

Step4. Polar representation of given complex number is:

$\therefore z=r\cos \theta +ir\sin \theta$

$=\sqrt{2}\cos \frac{3\pi }{4}+i\sqrt{2}\sin \frac{3\pi }{4}$

$=\sqrt{\mathbf{2}}\mathbf{}\mathbf{(}\mathbf{cos}\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{4}}\mathbf{}\mathbf{+}\mathbf{i}\mathbf{sin}\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{4}}\mathbf{}\mathbf{)}$

Hence, correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}$