Q.2 (b) Prove that \(\log\left(e^{i\alpha}+e^{i\beta}\right)=\log\left[2\cos\left(\frac{\alpha-\beta}{2}\right)\right]+i\left(\frac{\alpha+\beta}{2}\right)\).
Best answer
Take out \(e^{i(\alpha+\beta)/2}\):
\(e^{i\alpha}+e^{i\beta}=e^{i\frac{\alpha+\beta}{2}}\left(e^{i\frac{\alpha-\beta}{2}}+e^{-i\frac{\alpha-\beta}{2}}\right)=e^{i\frac{\alpha+\beta}{2}}\cdot2\cos\frac{\alpha-\beta}{2}\)
(using \(e^{i\theta}+e^{-i\theta}=2\cos\theta\)).
Taking logarithm (principal value, \(\cos\frac{\alpha-\beta}{2}>0\)) and using \(\log(ab)=\log a+\log b\), \(\log e^{i\phi}=i\phi\):
\(\log\left(e^{i\alpha}+e^{i\beta}\right)=\log\left[2\cos\frac{\alpha-\beta}{2}\right]+i\,\frac{\alpha+\beta}{2}\) (proved).
(Alternatively: \(e^{i\alpha}+e^{i\beta}=(\cos\alpha+\cos\beta)+i(\sin\alpha+\sin\beta)=2\cos\frac{\alpha-\beta}{2}\left[\cos\frac{\alpha+\beta}{2}+i\sin\frac{\alpha+\beta}{2}\right]\), giving modulus \(2\cos\frac{\alpha-\beta}{2}\) and amplitude \(\frac{\alpha+\beta}{2}\).)
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