Question

If $\frac{1-ix}{1+ix}=a+ib$, then $a^2+b^2$=
(a) 1
(b) −1
(c) 0
(d) none of these

Answer 1

(a) 1

$$\begin{gathered} \frac{1-ix}{1+ix}=a+ib \\ \mathrm{Taking}\mathrm{modulus}\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ \left|\frac{1-ix}{1+ix}\right|=\left|a+ib\right| \\ \Rightarrow \frac{\sqrt{1^2+x^2}}{\sqrt{1^2+x^2}}=\sqrt{a^2+b^2} \\ \Rightarrow \sqrt{a^2+b^2}=1 \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides},\mathrm{we}\mathrm{get}: \\ {a}^2+b^2=1 \end{gathered}$$