Question

Which of the following are correct?
I. Modulus of $\frac{1+ i }{1- i }$ is $1$ .
II. Argument of $\frac{1+ i }{1- i }$ is $\frac{\pi}{2}$
III. Modulus of $\frac{1}{1+i}$ is $\sqrt{2}$
IV. Argument of $\frac{1}{1+i}$ is $\frac{\pi}{4}$

• A. I and II are correct
• B. III and IV are correct
• C. I, II and III are correct
• D. All are correct

Answer 1

Answer: (A)

We have,
$\frac{1+i}{1-i}=\frac{1+i}{1-i} \times \frac{1+i}{1+i}=\frac{1-1+2 i}{1+1}=i=0+i$
Now, let us put $0= r \cos \theta, 1= r \sin \theta$
Squaring and adding;
$r ^{2}=1$, i.e, $r =1$
So, $cos \theta=0, sin \,\theta=1$
$ \therefore \theta=\frac{\pi}{2}$
Hence, the modulus of $\frac{1+ i }{1- i }$ is 1 and the argument is $\frac{\pi}{2}$.
Now, $\frac{1}{1+ i }=\frac{1- i }{(1+ i )(1- i )}=\frac{1- i }{1+1}=\frac{1}{2}-\frac{ i }{2}$
Let $\frac{1}{2}=r \cos \theta,-\frac{1}{2}=r \sin \theta$
Proceeding as $r=\frac{1}{\sqrt{2}} ; \cos \theta=\frac{1}{\sqrt{2}}, \sin \theta=\frac{-1}{\sqrt{2}}$
$\therefore \theta=\frac{-\pi}{4}$
$[\because \cos \theta>\,0$ and $\sin \theta<\,0$ is in IV quadrant ]
Hence, the modulus of $\frac{1}{1+ i }$ is $\frac{1}{\sqrt{2}}$ and the argument is $-\frac{\pi}{4}$