Question

Consider two complex numbers $\alpha$ and $\beta$ as $\alpha ={\left(\frac{a+bi}{a-bi}\right)}^2+{\left(\frac{a-bi}{a+bi}\right)}^2,$ where $a,b\in R$ and $\beta =\frac{z-1}{z+1},$ where $|z|=1.$ Then

• A. both $\alpha$ and $\beta$ are purely real
• B. both $\alpha$ and $\beta$ are purely imaginary
• C. $\alpha$ is purely real and $\beta$ is purely imaginary
• D. $\beta$ is purely real and $\alpha$ is purely imaginary

Answer 1

The correct option is C $\alpha$ is purely real and $\beta$ is purely imaginary

$\alpha ={\left(\frac{a+bi}{a-bi}\right)}^2+{\left(\frac{a-bi}{a+bi}\right)}^2$
$\overline{\alpha }=\overline{{\left(\frac{a+bi}{a-bi}\right)}^2}+\overline{{\left(\frac{a-bi}{a+bi}\right)}^2}$
$\Rightarrow \overline{\alpha }={\left(\frac{a-bi}{a+bi}\right)}^2+{\left(\frac{a+bi}{a-bi}\right)}^2=\alpha$
$\Rightarrow \overline{\alpha }=\alpha$
$\Rightarrow \alpha$ is purely real.

$z\overline{z}=|z{|}^2=1\beta =\frac{z-1}{z+1}\Rightarrow \beta =\frac{z-z\overline{z}}{z+z\overline{z}}\Rightarrow \beta =\frac{1-\overline{z}}{1+\overline{z}}\Rightarrow \overline{\beta }=\frac{1-z}{1+z}\Rightarrow \overline{\beta }=-\beta$
Hence, $\beta$ is purely imaginary.