Question

If tan(θ+iφ)=tanα+isecα
Prove that
1)${\displaystyle e^{2\phi }=\cot \left(\frac{\phi }{2}\right)}$
2) ${\displaystyle 2\theta =n\pi +\frac{\pi }{2}+\alpha }$

Answer 1

tan(θ+iφ) = tanα+isecα     ∴tan(θ-iφ)=tanα-isecα

∴tan2θ = tan[(θ+iφ)+(θ-iφ)

${\displaystyle =\frac{\tan (\theta +i\phi )+\tan (\theta -i\phi )}{1-\tan (\theta +i\phi )\tan (\theta -i\phi )}}$

${\displaystyle =\frac{\tan (\theta +i\sec \alpha )+\tan (\theta -i\sec \alpha )}{1-\tan (\theta +i\sec \alpha )\tan (\theta -i\sec \alpha )}}$

${\displaystyle =\frac{2\tan \alpha }{1-({\tan }^2\alpha +{\sec }^2\alpha )}=\frac{2\tan \alpha }{-2{\tan }^2\alpha }-\cot \alpha =\tan (\frac{\pi }{2}+\alpha )}$

${\displaystyle \therefore 2\theta =n\pi +\frac{\pi }{2}+\alpha .}$ (general value).

Again ${\displaystyle \tan \left(2i\phi \right)=\tan [(\theta +i\phi )-\left[(\theta -i\phi )\right]}$

${\displaystyle =\frac{\tan (\theta +i\sec \alpha )+\tan (\theta -i\sec \alpha )}{1+\tan (\theta +i\sec \alpha )\tan (\theta -i\sec \alpha )}}$

${\displaystyle \therefore i\tan 2\phi =\frac{2i\sec \alpha }{2{\sec }^2\alpha }=i\cos \alpha }$

${\displaystyle \therefore \tanh 2\phi =\cos \alpha }$

${\displaystyle \therefore 2\phi ={\tanh }^{-1}\cos \alpha \cos \alpha \frac{1}{2}\log \left[\frac{1+\cos \alpha }{1-\cos \alpha }\right]}$

${\displaystyle =\frac{1}{2}\log \left[\frac{2{\cos }^2\left(\frac{\alpha }{2}\right)}{2{\sin }^2\left(\frac{\alpha }{2}\right)}\right]\log \cot \left(\frac{\alpha }{2}\right)}$