Question

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

• A. $1$
• B. ${\alpha }^2$
• C. $-1$
• D. $-{\alpha }^2$

Answer 1

Answer: (A)

$A+iB=\frac{1-i\alpha }{1+i\alpha }$
$\Rightarrow A=iB=\frac{1+i\alpha }{1-i\alpha }$
$\Rightarrow (A+iB)(A-iB)$
$=\frac{(1-i\alpha )(1+i\alpha )}{(1+i\alpha )(1-i\alpha )}=1$
$\Rightarrow A^2+B^2=1$