Question

If θ is the amplitude of $\frac{a+ib}{a-ib}$, than tan θ =
(a) $\frac{2a}{a^2+b^2}$
(b) $\frac{2ab}{a^2-b^2}$
(c) $\frac{a^2-b^2}{a^2+b^2}$
(d) none of these

Answer 1

(b) $\frac{2ab}{a^2-b^2}$

$$\begin{gathered} z=\frac{a+ib}{a-ib}\times \frac{a+ib}{a+ib} \\ \Rightarrow z=\frac{a^2+i^2{b}^2+2abi}{a^2-i^2{b}^2} \\ \Rightarrow z=\frac{a^2-b^2+2abi}{a^2+b^2} \\ \Rightarrow z=\frac{a^2-b^2}{a^2+b^2}+i\frac{2ab}{a^2+b^2} \\ \Rightarrow \mathrm{Re}\left(z\right)=\frac{a^2-b^2}{a^2+b^2},\mathrm{Im}\left(z\right)=\frac{2ab}{a^2+b^2} \\ \tan \alpha =\left|\frac{\mathrm{Im}\left(z\right)}{\mathrm{Re}\left(z\right)}\right| \\ =\frac{2ab}{a^2-b^2} \\ \alpha ={\tan }^{-1}\left(\frac{2ab}{a^2-b^2}\right) \\ \mathrm{Since},z\mathrm{lies}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant}.\mathrm{Therefore}, \\ \arg \left(z\right)=\alpha ={\tan }^{-1}\left(\frac{2ab}{a^2-b^2}\right) \\ \tan \theta =\frac{2ab}{a^2-b^2} \end{gathered}$$