Question

If $\tan (\alpha +i\beta )=e^{i\theta };$ where $\alpha ,\beta \in R$, $\theta \ne (2n+1)\frac{\pi }{2},n\in Z$ and $i=\sqrt{-1}$, then

• A. $\alpha$ is an odd multiple of $\frac{\pi }{2}$
• B. $\alpha$ is an even multiple of $\frac{\pi }{2}$
• C. $\alpha$ is an odd multiple of $\frac{\pi }{4}$
• D. $\alpha$ is an even multiple of $\frac{\pi }{4}$

Answer 1

The correct option is C $\alpha$ is an odd multiple of $\frac{\pi }{4}$

$\tan (\alpha +i\beta )=e^{i\theta }$
$\Rightarrow \tan (\alpha +i\beta )=\cos \theta +i\sin \theta$
Taking complex conjugate, we get
$\tan (\alpha -i\beta )=\cos \theta -i\sin \theta$

$\tan 2\alpha =\tan \left[\right(\alpha +i\beta )+(\alpha -i\beta \left)\right]$

$\Rightarrow \tan 2\alpha =\frac{\tan (\alpha +i\beta )+\tan (\alpha -i\beta )}{1-\tan (\alpha +i\beta )\tan (\alpha -i\beta )}$

$\Rightarrow \frac{1}{\tan 2\alpha }=\frac{1-\tan (\alpha +i\beta )\tan (\alpha -i\beta )}{\tan (\alpha +i\beta )+\tan (\alpha -i\beta )}$

$\Rightarrow \cot 2\alpha =\frac{1-({\cos }^2\theta +{\sin }^2\theta )}{2\cos \theta }$

$\Rightarrow \cot 2\alpha =0[\because \theta \ne (2n+1\left)\frac{\pi }{2}\right]$

$\Rightarrow 2\alpha =\frac{(2m+1)\pi }{2},m\in Z$

$\Rightarrow \alpha =\frac{(2m+1)\pi }{4},m\in Z$