If \(\tan(\alpha+i\beta)=e^{i\theta}\), prove that \(\alpha=\frac{n\pi}{2}+\frac{\pi}{4}\) and \(\beta=\frac12\log\left[\tan\left(\frac\pi4+\frac\theta2\right)\right]\). (asked in MU Dec 2025)

Answer: \(\tan2\alpha\to\infty\) gives \(\alpha=\frac{n\pi}{2}+\frac\pi4\); \(\tanh2\beta=\sin\theta\) gives \(\beta=\frac12\log\tan\left(\frac\pi4+\frac\theta2\right)\).

Given \(\tan(\alpha+i\beta)=e^{i\theta}=\cos\theta+i\sin\theta\). Taking the conjugate: \(\tan(\alpha-i\beta)=e^{-i\theta}\).

Finding \(\alpha\):

\[\tan2\alpha=\tan[(\alpha+i\beta)+(\alpha-i\beta)]=\frac{e^{i\theta}+e^{-i\theta}}{1-e^{i\theta}e^{-i\theta}}=\frac{2\cos\theta}{1-1}=\infty.\]

So \(2\alpha=n\pi+\frac\pi2\), i.e. \(\alpha=\dfrac{n\pi}{2}+\dfrac\pi4\).

Finding \(\beta\):

\[\tan(2i\beta)=\tan[(\alpha+i\beta)-(\alpha-i\beta)]=\frac{e^{i\theta}-e^{-i\theta}}{1+e^{i\theta}e^{-i\theta}}=\frac{2i\sin\theta}{2}=i\sin\theta.\]

But \(\tan(2i\beta)=i\tanh2\beta\). Hence \(\tanh2\beta=\sin\theta\), so

\[2\beta=\tanh^{-1}(\sin\theta)=\frac12\log\frac{1+\sin\theta}{1-\sin\theta}.\]

Now \(\dfrac{1+\sin\theta}{1-\sin\theta}=\dfrac{\left(\cos\frac\theta2+\sin\frac\theta2\right)^2}{\left(\cos\frac\theta2-\sin\frac\theta2\right)^2}=\left(\dfrac{1+\tan\frac\theta2}{1-\tan\frac\theta2}\right)^2=\tan^2\left(\frac\pi4+\frac\theta2\right)\).

\(\therefore\ 2\beta=\frac12\cdot2\log\tan\left(\frac\pi4+\frac\theta2\right)\Rightarrow\ \beta=\dfrac12\log\left[\tan\left(\dfrac\pi4+\dfrac\theta2\right)\right]\). Hence proved.