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,
$\frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{1-1+2i}{1+1}=i=0+i$
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 $-\frac{\pi }{4}$