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We have $\log [e^{iα} + e^{iβ} ]=\log[(\cos α+i\sin α)+(\cos β+i\sin β)] \\ =\log [(\cos α+\cos β)+i(\sin α+\sin β)] $
In terms of real and imaginary parts it can be written as,
$=\dfrac 12 \log [(\cos α+\cos β)^2+(\sin α+ \sin β)^2 ]+i \tan^{-1} [\dfrac {\sin α+\sin β}{\cos α+ \cos β}] \\ =\dfrac 12 \log[\cos^2 α + 2\cos α.\cos β + \cos^2 β + \sin^2 α+2 \sin α.\sin β + \sin^2 β] + i\tan^{-1} [\dfrac {2\sin \dfrac {α+β}2.\cos \dfrac {α-β}2 }{2\cos \dfrac {α+β}2.\cos\dfrac{α-β}2}] \\ =\dfrac 12 \log[(\cos^2 α + \sin^2 α) + (\cos^2 β + \sin^2 β) + \cos α.\cos β + \sin α.\sin β]+i\tan^{-1} \tan(\dfrac {α+β}2) \\ = \dfrac 12 \log[2+2 \cos (α-β) ]+i (\dfrac {α+β}2) \\ = \dfrac 12 \log[1+ \cos(α-β) ] + i (\dfrac {α+β}2) \\ = \dfrac 12 \log4 \cos^2 (\dfrac {α-β}2) + i(\dfrac {α+β}2) \\ =\log2 \cos(\dfrac {α-β}2) + i(\dfrac {α+β}2)$
Hence proved.