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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.2012-05-09
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    I can't seem to find a question anywhere...2012-05-09
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    @Dylan: Are you referring to the application of the Residue Theorem?2012-05-09
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    It is rhetorical.2012-05-09
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    @ J.M. and copper, it's an apparent contradiction, $2\pi i = 0$.2012-05-09
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    A simply disconnected contradiction.2012-05-09
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    Read @DylanMoreland's comment above.2012-05-09
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    I still don't see my error, can someone clarify what theorem requires simply connected.2012-05-09
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    In the last sentence, how do you know that function has a primitive?2012-05-09
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    @timur: My professor says it does.2012-05-09
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    Wikipedia states Cauchy's Integral Theorem for simply connected domains, my book only requires the domain to be arc-wise connected, open and non-empty, which $\mathbb{C}-[-1,1]$ definitely is. My first paragraph means that $\frac{d}{dz}log(\frac{1+z}{1-z}) = \frac{dz}{1-z^2}$ is true on $\mathbb{C}-[-1,1]$.2012-05-09
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    What book are you using?2012-05-09
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    Wouldn't the two poles cancel, giving the residue $0$?2012-05-09
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    @copper.hat: Complex Analysis by Eberhard Freitag and Rolf Busam.2012-05-09
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    @timur: No they both evaluate to positive 1/2.2012-05-09
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    Isn't $\frac1{(1-z)(1+z)}=\frac1{2(z+1)}-\frac1{2(z-1)}$ ?2012-05-09
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    I get a residue of $0$.2012-05-09
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    I mean, showing that the residue is $0$ would be a way to prove that a primitive exists.2012-05-09
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    @timur: No $\frac{1}{(1-z)(1+z)} = \frac{1}{2(1-z)} + \frac{1}{2(1+z)}$.2012-05-09
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    Ok, what is the residue of $\frac1{1-z}$ ?2012-05-09
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    I hope that when all is said and done someone will summarize this comment thread as an answer.2012-05-09
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    $1$ I thought, and similarly for the other fraction I thought, but I must be understanding the residue theorem wrong because you're right they are the same and in your case they do cancel. Ok I see now that it's most likely $-1$, I guess I should be careful in the future to make sure my factors are of the form $(z-a)$ and not $(a-z)$.2012-05-09
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    Also, in your book, the versions of the Cauchy Integral Theorem are on triangles (the triangle interior must also lie in the domain), star-shaped sets, etc., etc., not on arc-wise connected domains.2012-05-09
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    This is not the residue theorem. It is a calculation of the residue from a Laurent series that we are dealing with. You just need to locate the coefficient of $(z-a)^{-1}$.2012-05-09
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    $\log(\frac{1-z}{1+z})$ is NOT a primitive of your function on the ENTIRE domain, to cover your domain you need to take two different branches of the logarithm. In particular, for any closed curve, the same branch of the log cannot be a primitive on the entire curve.2012-05-09
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    Actually I believe $log(\frac{1-z}{1+z})$ IS a primitive of my function on the given domain, however it is not the composition of two analytic functions in the usual way. Observe that $\int\frac{-2dz}{1-z^2}$ is zero along any piecewise smooth closed curve on our domain, hence it has a primitive in said domain. We define this primitive by $log(\frac{1-z}{1+z}) := \int\frac{-2dz}{1-z^2}$2012-05-09

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