Question:
All the points in the set \(S=\left\{\frac{\alpha+i}{\alpha-i}:\alpha\in R\right\}\) (\(i=\sqrt{-1}\)) lie on a :
All the points in the set \(S=\left\{\frac{\alpha+i}{\alpha-i}:\alpha\in R\right\}\) (\(i=\sqrt{-1}\)) lie on a :
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For real \(\alpha\), \(\alpha-i\) is the conjugate of \(\alpha+i\), so the modulus of the quotient is 1.
Updated On: Nov 5, 2026
- straight line whose slope is \(-1\)
- straight line whose slope is 1.
- circle whose radius is 1.
- circle whose radius is \(\sqrt2\).
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The Correct Option is C
Solution and Explanation
For real \(\alpha\), \(\alpha-i=\overline{\alpha+i}\).
So \(\left|\frac{\alpha+i}{\alpha-i}\right|=\frac{|\alpha+i|}{|\overline{\alpha+i}|}=1\).
Every point of \(S\) is at distance 1 from the origin, i.e. lies on the circle \(|z|=1\) of radius 1.
Answer: (3) circle whose radius is 1.
So \(\left|\frac{\alpha+i}{\alpha-i}\right|=\frac{|\alpha+i|}{|\overline{\alpha+i}|}=1\).
Every point of \(S\) is at distance 1 from the origin, i.e. lies on the circle \(|z|=1\) of radius 1.
Answer: (3) circle whose radius is 1.
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