Question

If $\alpha$ and $\beta$ are two distinct complex numbers satisfying $|\alpha {|}^2\beta -|{\beta }^2|\alpha =\alpha -\beta ,$ then
(Here, $\arg \left(z\right)$ denotes the principal argument with $-\pi <\arg \left(z\right)\le \pi$)

• A. $\arg \left(\frac{\alpha }{\beta }\right)=\pi$
• B. $\alpha \overline{\beta }=\beta \overline{\alpha }$
• C. $\alpha \overline{\beta }=1$
• D. $|\alpha |=|\beta |$

Answer 1

Answer: (B), (C)

$\alpha \overline{\beta }=1$

$|\alpha {|}^2\beta -|{\beta }^2|\alpha =\alpha -\beta \cdots \left(1\right)$
$\Rightarrow |\alpha {|}^2\beta +\beta =\alpha +|\beta {|}^2\alpha$
$\Rightarrow \beta (1+|\alpha {|}^2)=\alpha (1+|\beta {|}^2)$
$\Rightarrow \frac{\alpha }{\beta }=\frac{1+|\alpha {|}^2}{1+|\beta {|}^2}$
$\Rightarrow \frac{\alpha }{\beta }$ is positive real.
$\Rightarrow \arg \left(\frac{\alpha }{\beta }\right)=0$

Also, $\frac{\alpha }{\beta }=\frac{\overline{\alpha }}{\overline{\beta }}$
$\Rightarrow \alpha \overline{\beta }=\beta \overline{\alpha }$

Again, from $\left(1\right),\alpha \overline{\alpha }\beta -\beta \overline{\beta }\alpha =\alpha -\beta$
$\Rightarrow \alpha (\overline{\alpha }\beta -1)=\beta (\alpha \overline{\beta }-1)$
But $\overline{\alpha }\beta =\alpha \overline{\beta }$
Hence, $(\overline{\alpha }\beta -1)(\alpha -\beta )=0$
$\alpha \ne \beta \Rightarrow \overline{\alpha }\beta =1=\alpha \overline{\beta }$