If $\frac{1-i \alpha}{1+i \alpha}=A+i B$, then $A^2+B^2$ equals to

Question

If $\frac{1-i \alpha}{1+i \alpha}=A+i B$, then $A^2+B^2$ equals to, 1, $\alpha^2$, -1, $-\alpha^2$

MARKS App · Accepted Answer

$\begin{aligned} & A+i B=\frac{1-i \alpha}{1+i \alpha} \Rightarrow A-i B=\frac{1+i \alpha}{1-i \alpha} \ & \Rightarrow(A+i B)(A-i B)=\frac{(1-i \alpha)(1+i \alpha)}{(1+i \alpha)(1-i \alpha)}=1 \ & \Rightarrow A^2+B^2=1 \end{aligned}$

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