Question

All the points in $A=\{\frac{\lambda+i}{\lambda-i}; \lambda \in \mathbb{R}\}$ lie on ________.

• A. a circle with radius $\sqrt{2}$
• B. a circle with radius 2
• C. a circle with radius $\frac{1}{2}$
• D. a circle with radius 1
• E. a straight line with slope 1

Hint:

The modulus of $\frac{a+bi}{a-bi}$ is always 1.

Answer 1

The Correct Option is D

Step 1: Concept
Check the modulus of the complex number $z = \frac{\lambda+i}{\lambda-i}$.

Step 2: Meaning
$|z| = |\frac{\lambda+i}{\lambda-i}| = \frac{|\lambda+i|}{|\lambda-i|}$.

Step 3: Analysis
$|z| = \frac{\sqrt{\lambda^2+1^2}}{\sqrt{\lambda^2+(-1)^2}} = \frac{\sqrt{\lambda^2+1}}{\sqrt{\lambda^2+1}} = 1$.

Step 4: Conclusion
Since $|z|=1$, all points lie on a circle with radius 1 centered at the origin. Final Answer: (D)