Question

Consider two complex numbers $\alpha$ and $\beta$ as $\alpha =\left[\right(a+bi\left)/\right(a-bi){]}^2+[(a-bi)/(a+bi){]}^2$, where $a,bϵR$ and $\beta =(z-1)/(z+1)$, where $|z|=1$, then find the correct statement.

• 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 B Both $\alpha$ and $\beta$ are purely imaginary

$\alpha ={\left[\frac{a+ib}{a-ib}\right]}^2+{\left[\frac{a-ib}{a+ib}\right]}^2$

$\alpha ={\left[\frac{e^{i\theta }}{e^{-i\theta }}\right]}^2$$+{\left[\frac{e^{-i\theta }}{e^{i\theta }}\right]}^2$

$\alpha =e^{4i\theta }+e^{-4i\theta }$

$\alpha =ki⟹$ (purely imaginary)

$\beta =\frac{z-1}{z+1}$

componendo dividendo

$\frac{\beta +1}{\beta -1}=\frac{z-1+z+1}{z-1-z+1}\frac{\beta +1}{\beta -1}=\frac{-2z}{2}\frac{1+\beta }{1-\beta }=z$

Let $\beta =x+iy$

$\frac{(x+1)+iy}{(1-x)-iy}=z$, now $|z|=1$

$⟹\sqrt{(1+x{)}^2+y^2}$$=\sqrt{(1-x{)}^2+y^2}$

$⟹1+x=1-x$

$⟹x=0,y=1$

Again, $z=\frac{1+i}{1-i}=i$ (purely imaginary)