Show that $$ -\log(1-\mathrm{e}^{\mathrm{i}x}) = -\log\left(2\sin\left(\frac{x}{2}\right)\right) + \mathrm{i}\dfrac{\pi - x}{2}. $$ This is a last step in one of my problems, and I know the two expressions are equivalent, yet I can't seem to figure out how to manipulate the LHS to give me the RHS. Obviously I'd need to expand $e^{ix}$ using Euler's formula, but I can't see what they did. Any help would be greatly appreciated!
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actually, $\pi$ should be replaced by $2n\pi$ – Isura Manchanayake Nov 30 '15 at 09:35
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I see how you might figure that, but this suggests that what I put is also true. How can that be? – Nimmer Nov 30 '15 at 09:42
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Mh.. I think it is false. $\dfrac{1}{1-e^{ix}}=\dfrac{1}{1-e^{i(2n\pi+x)}}=\dfrac{1}{2\sin\dfrac{x}{2}}e^{-i\left(\dfrac{2n\pi+x}{2}\right)}=\dfrac{1}{2\sin\dfrac{x}{2}}e^{i\left(\dfrac{2m\pi-x}{2}\right)},n=-m$ – Isura Manchanayake Nov 30 '15 at 09:54
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Hmm. Let me provide the context of my problem. Restricting $0<x\leq \pi$ should give us our answer, correct? – Nimmer Nov 30 '15 at 10:04
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I see nothing false in my proof. I wonder how it has stated on two problems. Please comment here one day if you can surprise me. – Isura Manchanayake Nov 30 '15 at 19:24
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$1-e^{ix}=1-(cos\ x + i\ sin\ x) = 2\ sin^2\dfrac{x}{2} + 2\ i\ sin\dfrac{x}{2} cos\dfrac{x}{2} = 2\ sin\dfrac{x}{2}(cos\dfrac{x}{2}+i\ sin\dfrac{x}{2})=2\ sin\dfrac{x}{2}e^{\frac{ix}{2}}$
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Awesome, thanks. From here, I got $-ln(2sin(\frac{x}{2})) + i(\frac{x}{2})$. But $e^{\frac{ix}{2}} \neq e^{\frac{i(\pi - x)}{2}}$. Where does the $\pi$ come in? – Nimmer Nov 30 '15 at 09:36
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actually, $log\ z = log|z|+i(arg\ z + 2n\pi)$ Note $e^{ix}=cos\ x + i\ sin\ x \Rightarrow e^{ix}=e^{i(2n\pi+x)}$ – Isura Manchanayake Nov 30 '15 at 09:42
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I didn't include that my problem has x restricted so that $x \neq 2\pi k$ where $k$ is any integer. I assume that should change some things. – Nimmer Nov 30 '15 at 09:46