Question

All the possible points in the set $S=\{\frac{\alpha +i}{\alpha -i}:\alpha \in R,i=\sqrt{-1}\}$ lies on

• A. a circle whose radius is $\sqrt{2}$ units.
• B. a circle whose radius is $1$ unit.
• C. a straight line whose slope is $1.$
• D. a straight line whose slope is $-1.$

Answer 1

The correct option is B a circle whose radius is $1$ unit.

Let $\frac{\alpha +i}{\alpha -i}=Z$
Taking modulus on both sides
$\left|\frac{\alpha +i}{\alpha -i}\right|=\left|Z\right|$
As $\alpha$ is real,
$\therefore \frac{\sqrt{{\alpha }^2+1^2}}{\sqrt{{\alpha }^2+(-1{)}^2}}=\left|Z\right|\Rightarrow \left|Z\right|=1$

Hence, the points in $S$ lies on a circle with radius $1\text{unit}$.