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

If $a > 0$ and $z = \frac{{{{\left( {1 + i} \right)}^2}}}{{a - i}}$, has magnitude $\sqrt {\frac{2}{5}} $, then $\bar z$ is equal to:

If $z$ is a complex number such that $\frac{{z - 1}}{{z + 1}}$ is purely imaginary, then

$\left| {\frac{1}{2}({z_1} + {z_2}) + \sqrt {{z_1}{z_2}} } \right| + \left| {\frac{1}{2}({z_1} + {z_2}) - \sqrt {{z_1}{z_2}} } \right|$ =

Let ${z_1}$ be a complex number with $|{z_1}| = 1$ and ${z_2}$be any complex number, then $\left| {\frac{{{z_1} - {z_2}}}{{1 - {z_1}{{\bar z}_2}}}} \right| = $

If $\alpha $ and $\beta $ are different complex numbers with $|\beta | = 1$, then $\left| {\frac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right|$ is equal to