If z - \alpha \over z + \alpha ( \alpha \in R ) is a purely imaginary number and | z | = 2, then a value of \alpha is :

Complex Numbers PYQ — JEE Main 2019 · 12 Jan · Shift 1 · Q26 with Solution

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Correct answer: C

Let $z=x+iy,$ where $x,y\in \mathbb R$ and given $|z|=2\implies x^2+y^2=4.$
We are told that $\frac{z-\alpha}{z+\alpha}$ is purely imaginary, where $\alpha\in\mathbb R$ .

A complex number is purely imaginary iff its real part is $0$ . So let us compute the real part of

Rationalize the denominator: $\frac{x-\alpha+iy}{x+\alpha+iy}\cdot \frac{x+\alpha-iy}{x+\alpha-iy}.$

Then numerator becomes $((x-\alpha)+iy)((x+\alpha)-iy).$ Expand: $=(x-\alpha)(x+\alpha) + y^2 + i\big(y(x+\alpha)-y(x-\alpha)\big).$

Now, $(x-\alpha)(x+\alpha)=x^2-\alpha^2,$ and $y(x+\alpha)-y(x-\alpha)=2\alpha y.$

So numerator is $x^2+y^2-\alpha^2 + i(2\alpha y).$

Denominator is $(x+\alpha)^2+y^2.$

Hence,

\frac{x^2+y^2-\alpha^2}{(x+\alpha)^2+y^2} +i\frac{2\alpha y}{(x+\alpha)^2+y^2}.$$ 4. For this to be purely imaginary, its real part must be zero: $$\frac{x^2+y^2-\alpha^2}{(x+\alpha)^2+y^2}=0.$$ So, $$x^2+y^2-\alpha^2=0.$$ Using $x^2+y^2=|z|^2=4$, $$4-\alpha^2=0$$ $$\implies \alpha^2=4$$ $$\implies \alpha=\pm 2.$$ 5. From the options, the available value is $$\alpha=2.$$ So the correct option is **C**.