All the points in the set


All the points in the set $S=\left\{\frac{\alpha+i}{\alpha-i}: \alpha \in R\right\}(i=\sqrt{-1})$ lie on a:

  1. (1) straight line whose slope is 1 .

  2. (2) circle whose radius is 1 .


  3. (3) circle whose radius is $\sqrt{2}$.

  4. (4) straight line whose slope is $-1$.

Correct Option: , 2


Let $\mathrm{z} \in \mathrm{S}$ then $z=\frac{\alpha+i}{\alpha-i}$

Since, $z$ is a complex number and let $z=x+i y$

Then, $x+i y=\frac{(\alpha+i)^{2}}{\alpha^{2}+1}$ (by rationalisation)

$\Rightarrow x+i y=\frac{\left(\alpha^{2}-1\right)}{\alpha^{2}+1}+\frac{i(2 \alpha)}{\alpha^{2}+1}$

Then compare both sides

$x=\frac{\alpha^{2}-1}{\alpha^{2}+1}$ ...(1)

$y=\frac{2 \alpha}{\alpha^{2}-1}$........(2)

Now squaring and adding equations (1) and (2)

$\Rightarrow x^{2}+y^{2}=\frac{\left(\alpha^{2}-1\right)^{2}}{\left(\alpha^{2}+1\right)^{2}}+\frac{4 \alpha^{2}}{\left(\alpha^{2}+1\right)^{2}}=1$

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