Question:
If $\frac{1-i x}{1+i x}=a+i b$, then $a^{2}+b^{2}=$
(a) 1
(b) −1
(c) 0
(d) none of these
Solution:
(a) 1
$\frac{1-i x}{1+i x}=a+i b$
Taking modulus on both the sides, we get:
$\left|\frac{1-i x}{1+i x}\right|=|a+i b|$
$\Rightarrow \frac{\sqrt{1^{2}+x^{2}}}{\sqrt{1^{2}+x^{2}}}=\sqrt{a^{2}+b^{2}}$
$\Rightarrow \sqrt{a^{2}+b^{2}}=1$
Squaring both the sides, we get:
$a^{2}+b^{2}=1$
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