Question

All the points in the set $S = \begin{Bmatrix} \frac{\alpha + i}{\alpha - i} ; \alpha \in R (i = \sqrt{-1}) \text{ lie on a}\end{Bmatrix}$

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

Answer 1

Answer: (A)

circle whose radius is 1

Answer 2

Let $z = \frac{\alpha + i}{\alpha - i}$, where $\alpha \in \mathbb{R}$. We calculate the modulus of $z$: $|z| = \left| \frac{\alpha + i}{\alpha - i} \right| = \frac{|\alpha + i|}{|\alpha - i|}$ The modulus of $\alpha + i$ is $\sqrt{\alpha^2 + 1^2} = \sqrt{\alpha^2 + 1}$. The modulus of $\alpha - i$ is $\sqrt{\alpha^2 + (-1)^2} = \sqrt{\alpha^2 + 1}$. Therefore, $|z| = \frac{\sqrt{\alpha^2 + 1}}{\sqrt{\alpha^2 + 1}} = 1$. The condition $|z|=1$ signifies that the complex number $z$ lies on a circle centered at the origin with a radius of 1. We can also write $z = x+iy$. Then $|z| = \sqrt{x^2+y^2}$. So, $\sqrt{x^2+y^2} = 1$, which implies $x^2+y^2 = 1$. This is the equation of a circle centered at $(0,0)$ with radius 1. The question asks where all the points in the set $S$ lie. Since $|z|=1$ for all points in $S$, all these points lie on a circle whose radius is 1.