If frac{z - alpha}{z + alpha} (where alpha in | vedclass

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(A) Let $w = \frac{z - \alpha}{z + \alpha}$. Since $w$ is purely imaginary,$w + \bar{w} = 0$.$\frac{z - \alpha}{z + \alpha} + \overline{\left( \frac{z

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$2$

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(A) Let $w = \frac{z - \alpha}{z + \alpha}$. Since $w$ is purely imaginary,$w + \bar{w} = 0$.$\frac{z - \alpha}{z + \alpha} + \overline{\left( \frac{z - \alpha}{z + \alpha} \right)} = 0$$\frac{z - \alpha}{z + \alpha} + \frac{\bar{z} - \alpha}{\bar{z} + \alpha} = 0$$(z - \alpha)(\bar{z} + \alpha) + (\bar{z} - \alpha)(z + \alpha) = 0$$(z\bar{z} + z\alpha - \alpha\bar{z} - \alpha^2) + (z\bar{z} + \alpha\bar{z} - \alpha z - \alpha^2) = 0$$2z\bar{z} - 2\alpha^2 = 0$$|z|^2 = \alpha^2$Given $|z| = 2$,we have $2^2 = \alpha^2$,so $\alpha^2 = 4$.Thus,$\alpha = \pm 2$. The value $2$ is given in the options.

If $\frac{z - \alpha}{z + \alpha}$ (where $\alpha \in R$) is a purely imaginary number and $|z| = 2$,then a value of $\alpha$ is

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