Q.2 (c) If \(\tan(\alpha+i\beta)=e^{i\theta}\), prove that \(\alpha=\frac{n\pi}2+\frac\pi4\) and \(\beta=\frac12\log\left[\tan\left(\frac\pi4+\frac\theta2\right)\right]\).
Best answer
\(\tan(\alpha+i\beta)=\cos\theta+i\sin\theta\Rightarrow\tan(\alpha-i\beta)=\cos\theta-i\sin\theta\).
For \(\alpha\): \(\tan2\alpha=\tan[(\alpha+i\beta)+(\alpha-i\beta)]=\frac{2\cos\theta}{1-(\cos^2\theta+\sin^2\theta)}=\frac{2\cos\theta}{0}=\infty\).
\(\Rightarrow2\alpha=n\pi+\frac\pi2\Rightarrow\) \(\alpha=\frac{n\pi}2+\frac\pi4\).
For \(\beta\): \(\tan(2i\beta)=\tan[(\alpha+i\beta)-(\alpha-i\beta)]=\frac{2i\sin\theta}{1+1}=i\sin\theta\).
Since \(\tan(ix)=i\tanh x\): \(i\tanh2\beta=i\sin\theta\Rightarrow2\beta=\tanh^{-1}(\sin\theta)=\frac12\log\frac{1+\sin\theta}{1-\sin\theta}\).
\(\frac{1+\sin\theta}{1-\sin\theta}=\frac{\left(\cos\frac\theta2+\sin\frac\theta2\right)^2}{\left(\cos\frac\theta2-\sin\frac\theta2\right)^2}=\left(\frac{1+\tan\frac\theta2}{1-\tan\frac\theta2}\right)^2=\tan^2\left(\frac\pi4+\frac\theta2\right)\).
\(2\beta=\frac12\cdot2\log\tan\left(\frac\pi4+\frac\theta2\right)\Rightarrow\) \(\beta=\frac12\log\left[\tan\left(\frac\pi4+\frac\theta2\right)\right]\) (proved).
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