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Let , $$f(z)=\frac{1}{z}.\frac{1-2z}{z-2}\cdots \frac{1-10z}{z-10}$$Find $$\int_{|z|=100}f(z)\,dz$$

We find that the function $f(z)$ has simple pole at the points $z=0,2,4,6,8,10$ , and all the points lie in $|z|=100$. So the required integral equal to $2\pi i\times\text{sum of the residues}$. But that process is too much laborious in this case.

Does there any other simplest way to evaluate the integral ?

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  • $\left|\frac{1-az}{z-a}\right|=1$ on the unit circle. – A.Γ. Aug 08 '15 at 18:26
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    I'll leave this as a comment because I'm not very confident: Following @A.G. 's hint, a variation of the Schwarz reflection principle shows that $f$ satisfies $f(z)=1/\overline{f(1/\bar{z})}$ and the desired integral is equivalent to $$ \int_{|z|=1/100}\frac{1}{\overline{f(\bar{z})}},dz $$ The Schwarz reflection shows that $1/\overline{f(1/\bar{z})}$ is meromorphic, and precomposing with $1/z$ doesn't change this. Computing shows that it is holomorphic within the new contour. This (should) show that our desired integral is zero. – Blake Aug 08 '15 at 22:57
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    @Blake One more thing: when doing the inversion $1/z$ one has to recalculate $dz$ too, which becomes $d(1/z)=-dz/z^2$ and sets a pole at zero. But it is just one residue to calculate. – A.Γ. Aug 08 '15 at 23:01
  • @A.G. There I got forgetting how to substitute. Thanks – Blake Aug 08 '15 at 23:02
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    @Blake Here is a link on residue at infinity – A.Γ. Aug 08 '15 at 23:11
  • @Blake One more thing. The mapping reverses the orientation of the contour. So, watch out for that minus sign. – Mark Viola Aug 09 '15 at 00:08

1 Answers1

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Inasmuch as all of the singularities are contained within $|z|<100$, we can evaluate the contour integral by finding the Residue at Infinity. To do this, we evaluate

$$\text{Res}\left(f(z),z=\infty\right)=\text{Res}\left(-\frac{1}{z^2}f\left(\frac1z\right),z=0\right)$$

Now, given $f(z)=\frac1z\prod_{n=2,n even}^{10}\frac{1-nz}{z-n}$, we find that

$$\bbox[5px,border:2px solid #C0A000]{f\left(\frac1z\right)=z\prod_{n=2,n even}^{10}\frac{z-n}{1-nz}}$$

Therefore,

$$\text{Res}\left(f(z),z=\infty\right)=-\lim_{z\to 0}\left(\prod_{n=2,n even}^{10}\frac{z-n}{1-nz}\right)=\prod_{n=2,n even}^{10}\,n=3840$$

Finally, we have

$$\bbox[5px,border:2px solid #C0A000]{\oint_{|z|=100} f(z)\,dz=-2\pi i (3840)}$$

Mark Viola
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