Question

Statement I Polar form of $\frac{1+7i}{(2-i{)}^2}$ is $\sqrt{2}\left[\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right]$.
Statement II Polar form of $\frac{1+3i}{1-2i}$ 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 $a+ib$ form and then reduce them into polar form
I. Let $z=\frac{1+7i}{(2-i{)}^2}=\frac{1+7i}{4+i^2-4i}$
[use $(a-b{)}^2=a^2+b^2-2ab$ ]
$=\frac{1+7i}{4-1-4i}\left(\because i^2=-1\right)$
$=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}$
$=\frac{3+4i+21i+28i^2}{(3{)}^2-(4i{)}^2}\left[\because (a-b)(a+b)=a^2-b^2\right]$
$=\frac{3+i(4+21)-28}{9-16i^2}=\frac{-25+25i}{9+16}\left(\because i^2=-1\right)$
$\Rightarrow z=\frac{-75+75i}{25}=-1+i$
Now, let $-1+i=r\cos 0+ir\sin 0$
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 $\frac{1+7i}{(2-i{)}^2}$.
II. Let $z=\frac{1+3i}{1-2i}\times \frac{1+2i}{1+2i}=\frac{1+2i+3i+3i\times 2i}{1^2-(2i{)}^2}$
$\left[\because (a+b)(a-b)=a^2-b^2\right]$
$=\frac{1+5i+6i^2}{1-4i^2}$
$\left(\because i^2=-1\right)$
$=\frac{1+5i-6}{1+4}=\frac{-5+5i}{5}=-1+i,$