Question

Statement I Polar form of $\frac{1+7 i}{(2-i)^2}$ is $\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$.
Statement II Polar form of $\frac{1+3 i}{1-2 i}$ is $\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$

• A. Statement I is correct
• B. Statement II is correct
• C. Both are correct
• D. Neither I nor II is correct

Answer 1

Answer: (C)

First we will convert the given expressions into $a+i b$ form and then reduce them into polar form
I. Let $z=\frac{1+7 i}{(2-i)^2}=\frac{1+7 i}{4+i^2-4 i}$
[use $(a-b)^2=a^2+b^2-2 a b$ ]
$=\frac{1+7 i}{4-1-4 i} \left(\because i^2=-1\right)$
$=\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}$
$=\frac{3+4 i+21 i+28 i^2}{(3)^2-(4 i)^2}\left[\because(a-b)(a+b)=a^2-b^2\right]$
$=\frac{3+i(4+21)-28}{9-16 i^2}=\frac{-25+25 i}{9+16} \left(\because i^2=-1\right)$
$\Rightarrow z=\frac{-75+75 i}{25}=-1+i$
Now, let $-1+i=r \cos 0+i r \sin 0$
On comparing the real and imaginary parts of both sides, we get
$r \cos \theta =-1 $....(i)
and $ r \sin \theta =1 $...(ii)
Squaring and adding Eqs. (i) and (ii), we get
$r^2 \cos ^2 \theta+r^2 \sin ^2 \theta =(-1)^2+(1)^2 $
$r^2\left(\cos ^2 \theta+\sin ^2 \theta\right) =1+1$
$\Rightarrow r^2=2 \left(\because \cos ^2 \theta+\sin ^2 \theta=1\right)$
$\Rightarrow r=\sqrt{2}$
On dividing Eq. (ii) by Eq. (i), we get
$ \frac{r \sin \theta}{r \cos \theta}=\left|\frac{1}{-1}\right| \Rightarrow \tan \theta=1=\tan \frac{\pi}{4} $
$\Rightarrow \theta=\frac{\pi}{4}$
Since, the real part of $z$ is negative and the imaginary part of $z$ is positive. So, the point lies in II quadrant.
$\therefore \arg (z)=\pi-\theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$
$\therefore z=-1+i=\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$
which is the required polar form of $\frac{1+7 i}{(2-i)^2}$.
II. Let $z=\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}=\frac{1+2 i+3 i+3 i \times 2 i}{1^2-(2 i)^2}$
${\left[\because(a+b)(a-b)=a^2-b^2\right]}$
$ =\frac{1+5 i+6 i^2}{1-4 i^2} $
$ \left(\because i^2=-1\right) $
$ =\frac{1+5 i-6}{1+4}=\frac{-5+5 i}{5}=-1+i, $