Question

If $\theta$ is the amplitude of $\frac{a+ib}{a-ib},\text{then tan}\theta$

• 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

The correct option is B

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

$z=\frac{a+ib}{a-ib}\times \frac{a+ib}{a+ib}\Rightarrow z=\frac{a^2+i^2{b}^2+2abi}{a^2-i^2{b}^2}\Rightarrow z=\frac{a^2-b^2}{a^2+b^2}+i\frac{2ab}{a^2+b^2}\text{Re (z)}=\frac{a^2-b^2}{a^2+b^2},\text{Im (z)}=\frac{2ab}{a^2+b^2}tan\alpha =\left|\frac{Im\left(z\right)}{Re\left(z\right)}\right|=\frac{2ab}{a^2-b^2}\alpha =ta{n}^{-1}\left(\frac{2ab}{a^2-b^2}\right)$

Since, z lies in the first quadrant. Therefore,

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