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I was looking at this answer and I was confused to see the equality $$-\frac12\left[\;\log\left(1-e^{2ix}\right)+\log\left(1-e^{-2ix}\right)\;\right]=-\frac12\log(2-2\cos2x)$$ Which I am unable to prove.

I have tried $$ S=-\frac12\log(1-e^{2ix})-\frac12\log(1-e^{-2ix})=-\frac12\log\frac{1-e^{2ix}}{1-e^{-2ix}} $$ Then with $u=e^{ix}$: $$\frac{1-u^2}{1-\frac1{u^2}}=\frac{1-u^2}{1-\frac1{u^2}}\frac{u^2}{u^2}=-u^2\frac{u^2-1}{u^2-1}=-u^2$$ So $$S=-\frac12\log(-e^{2ix})=-\frac{i\pi}2-ix$$ Which I'm fairly certain is wrong. I'm sure there is something realy simple I'm missing here, could I have some help? Thanks.

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clathratus
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1 Answers1

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This step is wrong.

$$S=-\frac12\log(1-e^{2ix})-\frac12\log(1-e^{-2ix})=-\frac12\log\frac{1-e^{2ix}}{1-e^{-2ix}}$$

Correct way:

\begin{align} -\frac12[\log(1-e^{2ix})+\log(1-e^{-2ix})]&=-\frac12[\log((1-e^{2ix})(1-e^{-2ix}))]\\ &=-\frac12[\log(1-e^{-2ix}-e^{2ix}+1))]\\ &=-\frac12\log(2-2\cos2x) \end{align}

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