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Observe that: $\log(\frac{1+z}{1-z}) = -2\int\frac{dz}{1-z^2}$. (Not precisely true but read on)

Supposedly this function is analytic on the domain $\mathbb{C}-[-1,1]$, despite the fact that it's not decomposable into two analytic functions in the usual way, and thus shouldn't be thought of as a composition of functions.

Now for the closed curve $\alpha (t)=2e^{it}$ for $0\leq t\leq 2\pi$, $\int_{\alpha}\frac{-2dz}{1-z^2} = \int_{\alpha}\frac{-2dz}{(1-z)(1+z)} = 2\pi i$ by the Residue Theorem. Yet by the Cauchy Integral Theorem $\int\frac{-2dz}{1-z^2}$ having a primitive on $\mathbb{C}-[-1,1]$ implies that $\int_{\alpha}\frac{-2dz}{1-z^2} = 0$ for any closed curve on $\mathbb{C}-[-1,1]$.

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That set is not simply connected. –  Dylan Moreland May 9 '12 at 5:04
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I can't seem to find a question anywhere... –  J. M. May 9 '12 at 5:05
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Read @DylanMoreland's comment above. –  copper.hat May 9 '12 at 5:09
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@timur: My professor says it does. –  Ron Jeremy May 9 '12 at 5:22
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I hope that when all is said and done someone will summarize this comment thread as an answer. –  Gerry Myerson May 9 '12 at 5:53
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1 Answer 1

up vote 4 down vote accepted

The residue is $0$, since $$ \frac1{(1-z)(1+z)} = \frac12\left(\frac1{z+1}-\frac1{z-1}\right). $$

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