If θ is the amplitude of


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,

$\arg (z)=\alpha=\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)$

$\tan \theta=\frac{2 a b}{a^{2}-b^{2}}$

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