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Question: Show that for R > 1 $$\int_{|z|=1} \frac{z^{2011}}{2z^{2012}-1} dz = \int_{|z|=R} \frac{z^{2011}}{2z^{2012}-1} dz$$

Thoughts thus far: (i) I know that we cannot use Cauchy's integral formula because neither is f analytic (because there are singularities within the unit circle) nor would the reduced form fit the form of Cuachy's integral formula (i.e. z0 would not be fixed). (ii) The above integral may be evaluated to ln(2z^2012-1)/4024 using u substitution where u=2z^2012, but I am not certain that this evaluation is even correct. (iii) I have also considered reducing it to polar form and subtracting the two integrals to show that the difference is zero, but I get stuck with $$\int_{|z|=1} \frac{z'(\theta)z(\theta)^{2011}}{2z(\theta)^{2012}-1}dz - \int_{|z|=R} \frac{z'(\theta)z(\theta)^{2011}}{2z(\theta)^{2012}-1} d\theta$$ $$\int_{unit\space circle} \frac{i\theta e^{i\theta}e^{2011i\theta}}{2e^{2012i\theta}-1} - \frac{i\theta Re^{i\theta}R^{2011}e^{2011i\theta}}{2R^{2012}e^{2012i\theta}-1}\ d\theta$$ I am unable to deal with the R terms, which do not appear to disappear easily. I have also considered multiplying by the conjugate of z, but again we get stuck with R terms on the RHS. (iv) I am led to believe that, rather than simple ignorance of calculating the above integral, there is something conceptual or extremely fundamental that I am missing that allows us to disregard the value of R (for R > 1).

Any help would be greatly appreciated. Thank you in advance.

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You're making this much harder than it needs to be. Apply Cauchy's Residue Theorem. See my post below. –  Fly by Night Sep 30 '12 at 16:07
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This seems to be a straight forward application of Cauchy's Residue Theorem. It tells us that the integral of a function around a contour is $2\pi i$ times the sum of the residues at the poles inside the contour of the function.

Your function has a pole when $2z^{2012}-1=0$. There are 2,012 simple poles and they all lie on the circle $|z| = \sqrt[2012]{1/2} \approx 0.9997.$ Any circular contour $|z| = R$ will contain all of these poles in its interior provided $R > \sqrt[2012]{1/2}$. It follows that the integral around any circular contour $|z| = R$ will be the same provided $R > \sqrt[2012]{1/2}$. In your case $|z|=1$ and $|z|=R$ with $R>1$ are two such contours. (If $R < \sqrt[2012]{1/2}$ then the integral will be zero because the integrand is holomorphic on and inside the contour $|z| = R$.)

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The last statement is not clear: the integral is 0 even if the circle meets the roots (now with $r=\sqrt[2012]{1/2}$)? –  Berci Sep 30 '12 at 16:12
    
Thank you. I am unfamiliar with this theorem. In class we just covered Cauchy's integral formula. Would there be any simple way to sum up the residues given that there are 2012 of them? –  ABC Bach Sep 30 '12 at 16:21
    
@ABCBach You don't need to know the integral to prove the statement that you've been asked to prove. If you want to calculate the actual integral then you could work out all of the residues, but that would be very messy. Your best bet would be to integrate by substitution. But the question doesn't ask for the value of the integral, so there's no point supplying more information than is asked for. –  Fly by Night Sep 30 '12 at 16:32
    
@Berci The statement is perfectly clear. The integrand is holomorphic on and inside the contour $|z| = R$ if and only if $R < \sqrt[2012]{1/2}$. By Cauchy's Residue Theorem, the integral around the contour $|z|=R$ is zero if and only if $R < \sqrt[2012]{1/2}$. –  Fly by Night Sep 30 '12 at 16:33
    
May this result of Cauchy's Residue Theorem be generalized? That is, is this radius-independence for the integral of f(z)dz around a disc characteristic of all functions where r>R such that R is the radius of the disc containing the simple poles of f(z)? –  ABC Bach Sep 30 '12 at 16:51
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Hint: Draw a segment between the 2 circles, and perform 2 integrals over the given paths.

The evaluation with $\log$ is also correct. But for that you should now that $\log$ is not unique (since $\exp$ is periodic with $2\pi i$ period!), but unique up to $+2k\pi i$ for $k\in\mathbb Z$. If the origo (the pole) is surrounded only once, you get $$\int_{|z|=e^r}\frac 1z = [\log z]_{z=e^r}^{e^{r+2\pi i}} = (r+2\pi i)-r = 2\pi i$$

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Thank you for your response. Among easiest ways to connect these two pieces would be a line along the real axis. How would this integral look? Would it simply be $$\int_{|z|=1} \frac{z^{2011}}{2z^{2012}-1} dz + \int_1^R 1+z dz \int_{|z|=R} \frac{z^{2011}}{2z^{2012}-1} dz$$ –  ABC Bach Sep 30 '12 at 16:19
    
No. Integral the same function ($f(z)=\frac{z^{2011}}{2z^{2012}-1}$), and is analytic on both cycle paths. Once you integrate it to the right $1\to R$, the other time it goes to the left $R\to 1$.. use $\int_{\alpha+\beta} f = \int_\alpha f + \int_\beta f$ where $\alpha,\beta$ are directed paths in $\mathbb C$. –  Berci Sep 30 '12 at 16:22
    
So, if I am understanding this correctly, we are looking at the area $$\int_{|z|=R} \frac{z^{2011}}{2z^{2012}-1} dz - \int_{|z|=1} \frac{z^{2011}}{2z^{2012}-1} dz$$ since this would be analytic. I apologize if I still am not understanding it –  ABC Bach Sep 30 '12 at 16:29
    
I know this is a little late but Berci had a correct solution (in fact this was the solution that the teacher showed in class), and it is quite elegant. I am posting this solution for anyone else who may be encountering a similar problem. –  ABC Bach Oct 14 '12 at 7:03
    
We know that $\int_{|z|=R} \frac{z^{2011}}{2z^{2012}-1} dz = \int_{|z|=1} \frac{z^{2011}}{2z^{2012}-1} dz \iff \int_{|z|=R} \frac{z^{2011}}{2z^{2012}-1} dz - \int_{|z|=1} \frac{z^{2011}}{2z^{2012}-1} dz = \int_{|z|=1} \frac{z^{2011}}{2z^{2012}-1} dz - \int_{|z|=R} \frac{z^{2011}}{2z^{2012}-1} dz = 0$. We may consider the above integrals as a keyhole structure with a width $\gamma$ as $\gamma$ approaches 0. We know that the outlined area defines an analytic function since the singularities are located within the unit circle. –  ABC Bach Oct 14 '12 at 7:11
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This is not an answer to your original question and I think that has been already taken care of. But there is an alternative approach to evaluating the integral instead of computing all the residues. Note that when $z \in \{|z|=R\}$ with $R>>1$, we can write the integrand as

$\dfrac{z^{2011}}{2z^{2012}}(1-\dfrac{1}{2z^{2012}})^{-1} = \dfrac{1}{2z}\sum_{n=0}^{\infty}(\dfrac{1}{2z^{2012}})^n$.

When you integrate this function over the unit circle only the term involving $\frac{1}{z}$ will produce a non-zero residue and which will happen only for $n=0$ in the power series above. Hence the integral will be equal to $\pi i$.

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