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Let $\log$ be the branch of the logarithm that extends the usual real logarithm, and consider on $D=\Bbb C\smallsetminus [-1,1]$ the function $$f(z)=\log\frac{z+1}{z-1}$$

I have to find the integral of $f$ around the circle $|z|=2$. Now, as an example, consider the integral $$\int_{|z|=2}\frac{e^{z+z^{-1}}}{1-z^2}dz$$

Using the biholomorphic mapping $B(0,1)^\times\to D$ that sends $z\to \frac 1 2(z+z^{-1})$ I got the integral $$\frac 1 2\int_\gamma \frac{e^{2z}}{1-z^2}dz$$

and $\gamma$ is a closed path inside $B(0,1)^{\times}$. This means the integral vanishes. I am trying to do something similar here. So if I take $w=\frac{z+1}{z-1}$, I get that $\frac{(w-1)^2}2=\frac{2}{(z-1)^2}$ and $dw=-2dz/(z-1)^2$, so I ultimately want to look at $$-2\int_{\gamma^{-}}\frac{ \log w}{(w-1)^2}dw$$ where $\gamma$ is a circle that passes through $1/3,3,-3/5+4/5 i$. Using the computer I got the circle is $$(x-5/3)^2+(y-5/2)^2=\frac{7225}{900}$$

Add This circle's orientation is now reversed.

Even without this one sees $1$ is an interior point, so the integral should equal, by Cauchy, $4\pi i$ (not $-4\pi i$). Can anyone confirm this is correct, and/or suggest another approach?

Pedro
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  • $\int_{|z|=2}\log\frac{z+1}{z-1}dz=\int_{a=0}^{2\pi}(\log(2e^{ia}+1)-\log(2e^{ia}-1))2ie^{ia}da$. Mathematica showed that the integration is equal to $4\pi i$ – mike Nov 20 '14 at 07:47
  • @mike It might be the orientation of the circle gets reversed, and I missed that. – Pedro Nov 20 '14 at 08:47
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    Yes, @Pedro, the orientation gets reversed. Consider in order the points $2,2i,-2$ on $\lvert z\rvert = 2$. They are mapped to $$3,;\frac{1+2i}{-1+2i} = -\frac{(1+2i)^2}{5} = \frac{3-4i}{5},; \frac{1}{3}$$ in order, which means $\gamma$ is traversed clockwise. – Daniel Fischer Nov 20 '14 at 10:45

1 Answers1

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As mike said in the comments above, I think the answer is $4 \pi i$.

Since $1+ \frac{1}{z}$ and $1- \frac{1}{z}$ both lie in the right half-plane for $|z| >1$,

$$- \pi < \text{Arg} \left(1 + \frac{1}{z} \right) - \text{Arg} \left(1- \frac{1}{z} \right) \le \pi \ \ \text{for} \ \ |z| >1.$$

Thus

$$ \begin{align} f(z) = \log \left(\frac{z+1}{z-1} \right) &= \log \left( \frac{1 + \frac{1}{z}}{1- \frac{1}{z}} \right) \\&= \log \left(1 + \frac{1}{z} \right) - \log \left(1 - \frac{1}{z} \right) \\ &= \frac{2}{z} + \text{O}(z^{-3}) \ \ \text{for} \ |z| >1. \end{align}$$

So $$ \int_{|z|=2} \log \left(\frac{z+1}{z-1} \right) \ dz = - 2 \pi i \ \text{Res} [f(z), \infty] = - 2 \pi i (-2) = 4 \pi i $$ since by definition the residue at infinity of $f(z)$ is the negative of the coefficient of the $\frac{1}{z}$ term in the Laurent expansion at $\infty$.

Random Variable
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  • It is not clear to me how you derive the third equality. – Pedro Nov 20 '14 at 08:33
  • @PedroTamaroff $\log \left( \frac{1 + \frac{1}{z}}{1- \frac{1}{z}} \right) = \log \left( 1+ \frac{1}{z} \right) - \log \left(1 - \frac{1}{z}\right) $ and then expand both functions in Taylor series – Random Variable Nov 20 '14 at 08:40
  • The equation $\log(wz)=\log w+\log z$ doesn't always hold. Why does it hold in this case? – Pedro Nov 20 '14 at 08:43
  • @PedroTamaroff It holds because as Daniel Fischer pointed out to me, $1- \frac{1}{z}$ and $1 + \frac{1}{z}$ lie in the right half-plane for $|z| >1$. I would just let $z=x+iy$ to see that's the case. Thus $ - \pi < \text{Arg} \left(1 + \frac{1}{z} \right) - \text{Arg} \left(1- \frac{1}{z} \right) \le \pi $. – Random Variable Nov 20 '14 at 10:15
  • Also, suppose $\lvert z\rvert$ large (one could check whether $2$ is large enough, but by Cauchy's integral theorem we know we can take $\lvert z\rvert$ as large as we please, so why bother?), then we can use the Taylor series of $\log (1+w)$ on $$\frac{1+\frac{1}{z}}{1-\frac{1}{z}} = 1 + \frac{2}{z} + \frac{2}{z^2} + \dotsc.$$ – Daniel Fischer Nov 20 '14 at 10:37