Question

Let $z=x+iy$ be a complex number such that $|z|=1$, where $i=\sqrt{-1}$. Match List - I with List - II.

$\begin{array}{cccc}\text{List-I}& & \text{List - II}& \\ \text{(I)}& \text{Re}\left(\frac{iz}{1+z^2}\right)\text{is equal to}& \text{(P)}& 0\\ \text{(II)}& \text{Im}\left(\frac{iz}{1+z^2}\right)\text{can be equal to}& \text{(Q)}& 1\\ \text{(III)}& \text{Number of integers NOT in the}& \text{(R)}& \frac{1}{2}\\ & \text{range of}\text{Im}\left(\frac{iz}{1+z^2}\right)\text{is equal to}& & \\ \text{(IV)}& \frac{1}{2\pi }\arg \left(\frac{iz}{1+z^2}\right)\text{is equal to}& \text{(S)}& -\frac{1}{2}\\ & (\text{where}-\pi <\arg (z)\le \pi )& & \\ & & \text{(T)}& -\frac{1}{4}\\ & & \text{(U)}& \frac{1}{4}\end{array}$

Which of the following is only CORRECT combination?

• A. $\text{III}\to \text{P, Q}$
• B. $\text{IV}\to \text{T, U}$
• C. $\text{III}\to \text{S, T, U}$
• D. $\text{III}\to \text{P, Q, T}$

Answer 1

The correct option is B $\text{IV}\to \text{T, U}$

$|z|=1\Rightarrow \frac{1}{z}=\overline{z}$

Now $\frac{iz}{1+z^2}=\frac{i}{z+\frac{1}{z}}=\frac{i}{z+\overline{z}}$

$=\frac{i}{2\cos \theta }=\left(\frac{1}{2}\sec \theta \right)i$

$\Rightarrow \text{Re}\left(\frac{iz}{1+z^2}\right)=0$ and $\text{Im}\left(\frac{iz}{1+z^2}\right)=\frac{1}{2}sec\theta$

So, the range of imaginary part $\frac{iz}{1+z^2}$ is $(-\infty ,-\frac{1}{2}]\cup [\frac{1}{2},\infty )$

$\because \frac{iz}{1+z^2}$ is purely imaginary

$\arg \left(\frac{iz}{1+z^2}\right)=\frac{\pi }{2}$ or $-\frac{\pi }{2}$