0
3.5kviews
show that $\log[e^{i\alpha}+e^{i\beta}] =\log[2\cos(\dfrac {\alpha-\beta}2) + i(\dfrac {\alpha+\beta}2)]$
1 Answer
2
375views

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,

$=\dfrac 12 \log [(\cos α+\cos β)^2+(\sin α+ \sin β)^2 ]+i \tan^{-1} [\dfrac {\sin α+\sin β}{\cos α+ \cos β}] \\ =\dfrac 12 \log[\cos^2 α + 2\cos α.\cos β + \cos^2 β + \sin^2 α+2 \sin α.\sin β + \sin^2 β] + i\tan^{-1} [\dfrac {2\sin \dfrac {α+β}2.\cos \dfrac {α-β}2 }{2\cos \dfrac {α+β}2.\cos\dfrac{α-β}2}] \\ =\dfrac 12 \log⁡[(\cos^2 α + \sin^2 α) + (\cos^2 β + \sin^2 β) + \cos α.\cos β + \sin α.\sin β]+i\tan^{-1} \tan(\dfrac {α+β}2) \\ = \dfrac 12 \log⁡[2+2 \cos⁡ (α-β) ]+i (\dfrac {α+β}2) \\ = \dfrac 12 \log⁡[1+ \cos⁡(α-β) ] + i (\dfrac {α+β}2) \\ = \dfrac 12 \log4 \cos^2⁡ (\dfrac {α-β}2) + i(\dfrac {α+β}2) \\ =\log⁡2 \cos(\dfrac {α-β}2) + i(\dfrac {α+β}2)$

Hence proved.

Please log in to add an answer.