(D) Let $z = \frac{\alpha + i}{\alpha - i}$.Taking the modulus on both sides,we get $|z| = \left| \frac{\alpha + i}{\alpha - i} \right|$.Since $|\alph

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(D) Let $z = \frac{\alpha + i}{\alpha - i}$.Taking the modulus on both sides,we get $|z| = \left| \frac{\alpha + i}{\alpha - i} \right|$.Since $|\alpha + i| = \sqrt{\alpha^2 + 1}$ and $|\alpha - i| = \sqrt{\alpha^2 + (-1)^2} = \sqrt{\alpha^2 + 1}$,we have $|z| = \frac{\sqrt{\alpha^2 + 1}}{\sqrt{\alpha^2 + 1}} = 1$.The equation $|z| = 1$ represents a circle in the complex plane with center at the origin $(0, 0)$ and radius $1$.

Class 11 Mathematics 4-1.Complex numbers Geometry of complex numbers

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

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4-1.Complex numbers

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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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