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 $a,bϵR$ 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

Answer: (C)

$\alpha$ is purely real and $\beta$ is purely imaginary

$\alpha ={\left(\frac{a+bi}{a-bi}\times \frac{a+bi}{a+bi}\right)}^2+{\left(\frac{a-bi}{a+bi}\times \frac{a-bi}{a-bi}\right)}^2$

$=\frac{{\left(a+bi\right)}^4}{{\left(a^2+b^2\right)}^2}+\frac{{\left(a-bi\right)}^4}{{\left(a^2+b^2\right)}^2}$

$=\frac{1}{{\left(a^2+b^2\right)}^2}\left[{\left(a+bi\right)}^4+{\left(a-bi\right)}^4\right]$

$=\frac{1}{{\left(a^2+b^2\right)}^2}\left[2{\left(a^2-b^2\right)}^2-8a^2{b}^2\right]$

$\Rightarrow \alpha$ is purely real

Now $\beta =\frac{\left(x+iy\right)-1}{x+iy+1}$

$=\frac{\left[\left(x-1\right)+iy\right]}{\left[\left(x+1\right)+iy\right]}\times \frac{\left[\left(x+1\right)-iy\right]}{\left[\left(x+1\right)-iy\right]}$

$=\frac{\left[\left(x-1\right)+iy\right]\times \left[\left(x+1\right)-iy\right]}{{\left(x+1\right)}^2+y^2}$

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

$=\frac{2iy}{{\left(x+1\right)}^2+y^2}$

$\Rightarrow \beta$ is purely imaginary.