Question:
All the points in the set
$\mathrm{S}=\left\{\frac{\alpha+\mathrm{i}}{\alpha-\mathrm{i}}: \alpha \in \mathrm{R}\right\}(\mathrm{i}=\sqrt{-1})$ lie on a
Correct Option: 1
Solution:
Let $\frac{\alpha+\mathrm{i}}{\alpha-\mathrm{i}}=\mathrm{z}$
$\Rightarrow \frac{|\alpha+i|}{|\alpha-i|}=|z|$
$\Rightarrow 1=|z|$
$\Rightarrow$ circle of radius 1
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