I'm still a bit insecure when it comes to complex analysis and I wondered if you guys could take a look at my solution to this problem.

Let $a > 0 $ and define $$f(z) = \frac{\log(z)}{z^2 +a^2}$$ with $\log(z) = \log|z| + i \arg(z)$, with $-\pi < \arg z < \pi$.

Calculate $$\oint_\gamma f(z) dz$$ Where $\gamma$ is the (counterclockwise) contour of the following figure: The half circle with radius $R$ centered at the origin in the upper half plane without the half circle with radius $\epsilon$ in the upper half plane. It also holds that $0 < \epsilon < a <R$.

This is my solution: First I'm going to split up $f$: $$f(z) = \frac{\log(z) \frac{1}{ia}}{z-ia} + \frac{\log(z) \frac{-1}{ia}}{z+ia}.$$

I now expect a simple poles at $\pm ia$, since there obviously are isolated singularities at these points. We can easily verify this: $$ \lim_{z \to ia} (z-ia)f(z) = \lim_{z \to ia} \left( \log(z)\frac{1}{ia} + \frac{\log(z)(z-ia) \frac{-1}{ia}}{z+ia} \right) = \frac{\log{ia}}{ia}.$$

Since we have calculated the residue let's apply the residue theorem! $$\oint_\gamma f(z)dz = 2\pi i \left( \text{wind}(\gamma, ia) \text{Res}(f,ia)+ \text{wind}(\gamma, -ia) \text{Res}(f,-ia) \right).$$

Since $\text{wind}(\gamma, -ia) = 0$ we obtain: $$\oint_\gamma f(z)dz = 2 \pi i \left( \frac{log(ia)}{ia} \right).$$

Thanks in advance!l

  • $\begingroup$ This should also have different values depending on the radius of the two half circles. $\endgroup$ – ClassicStyle May 30 '14 at 15:13
  • $\begingroup$ Could you please elaborate a bit? Maybe try an answer $\endgroup$ – Leo May 30 '14 at 15:18
  • $\begingroup$ Think about how the half washer is defined. It doesn't necessarily have to contain any poles. What are the consequences of this? $\endgroup$ – ClassicStyle May 30 '14 at 15:59

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