If $\theta$ is the amplitude of $\frac{a+i b}{a-i b}$, than $\tan \theta=$
(a) $\frac{2 a}{a^{2}+b^{2}}$
(b) $\frac{2 a b}{a^{2}-b^{2}}$
(c) $\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$
(d) none of these
(b) $\frac{2 a b}{a^{2}-b^{2}}$
$z=\frac{a+i b}{a-i b} \times \frac{a+i b}{a+i b}$
$\Rightarrow z=\frac{a^{2}+i^{2} b^{2}+2 a b i}{a^{2}-i^{2} b^{2}}$
$\Rightarrow z=\frac{a^{2}-b^{2}+2 a b i}{a^{2}+b^{2}}$
$\Rightarrow z=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}+i \frac{2 a b}{a^{2}+b^{2}}$
$\Rightarrow \operatorname{Re}(z)=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}, \operatorname{Im}(z)=\frac{2 a b}{a^{2}+b^{2}}$
$\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$
$=\frac{2 a b}{a^{2}-b^{2}}$
$\alpha=\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)$
Since, $z$ lies in the first quadrant. Therefore,
$\arg (z)=\alpha=\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)$
$\tan \theta=\frac{2 a b}{a^{2}-b^{2}}$