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I need to check something that I think I got wrong at the first try. If $$ f(z) = \frac{1+z^2}{1-\cos{2\pi z}} $$ Find $$ I = \frac{1}{2\pi i } \oint_C \frac{f'(z)}{f(z)} dz $$ where $ C$ is the circle around $|z| = \pi $

I used the argument theorem, $$ \frac{1}{2\pi i }\oint_C \frac{f'(z)}{f(z)} dz = N - P $$ where $N$ is the sum of the order of the zeros, and $P$ is the sum of the order of poles inside the countour.

Zeros of $ f(z)$ are $ i, -i$, both or order $1$, then, the poles of $f(z)$ are $-3,-2,-1,0,1,2,3$. While doing my exam, I thought they were orden $1$ each one, so the integral would be $2-7 = -5$. But I think now, If I expand cosine, I would get $$\frac{1+z^2}{1- (1-\frac{z^2}{2!} +\frac{z^4}{4!} + \cdots )} =\frac{1+z^2}{ \frac{z^2}{2!} -\frac{z^4}{4!} + \cdots } $$ So, the pole should be now order $2$, so the integral should be $2 - 14 = -12$ $$I = -12$$ Is it right now or is there anything else I'm missing?

Kaliche
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    Yes, the poles are of order 2 since the first nonzero coefficient in the Laurent series is $c_{-2}$. Another way to see this is to notice that $\frac{1}{f(x)}$ has zeroes of order $2$ at the integers. So the contour integral indeed evaluates to $-12$ by the argument theorem/principle. – Random Variable Jun 10 '13 at 05:08
  • Here is how to find the order of a pole. – Mhenni Benghorbal Jun 11 '13 at 01:10

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This is a very nice exercise and it shows that one must try to be careful with orders of poles/zeros. We can see that if we put

$$f(z)=\frac{z^2+1}{1-\cos 2\pi z}$$

then, for example:

$$\lim_{z\to 0}zf(z)\stackrel{\text{l'Hospital}}=\lim_{z\to 0}\frac{3z^2+1}{2\pi\sin 2\pi z}\;\;--\text{ doesn't exist finitely , but}$$

$$\lim_{z\to 0}z^2f(z)\stackrel{\text{l'Hospital}}=\lim_{z\to 0}\frac{4z^3+2z}{2\pi\sin 2\pi z}\stackrel{\text{l'Hospital}}=\lim_{z\to 0}\frac{12z^2+2}{2\pi^2\cos2\pi z}=\frac1{\pi^2}$$

so that the pole $\;z=0\;$ is a double one, and something similar happens with the other polesof the function.

DonAntonio
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