Computing $\int_{|z|=2} z^n(1 - z)^m\ dz$
I need help for this question when $m, n$ are negative integers.
Thanks.
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Apply the residue theorem... – Rushabh Mehta Mar 14 '20 at 23:49
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1Until now we did not take this theorem....so far we take Cauchy integral formula – ks1 Mar 14 '20 at 23:52
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How about Laurent series? Compute the Laurent series valid in $|z|>1$, then integrate term-by-term. – GEdgar Mar 14 '20 at 23:57
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The answer depends on the sum $n+m$. If $n+m<-1$ Then the integral is zero. \begin{gather} f(z)=z^n(1-z)^m\\ I=-2\pi i\underset{z=\infty}{\rm res}f(z)=2\pi i c_{-1} \end{gather} There are no poles outside the contour. Here, $c_{-1}$ is the coefficient of $f(z)$ expansion in the vicinity of the infinity. The leading behavior of $f(z)$ is \begin{gather} f(z)=(-1)^mz^{n+m}\ \ z\rightarrow\infty. \end{gather} Therefore $c_{-1}=0$ unless $m+n=-1$.
If $m+n=-1$ then: \begin{gather} c_{-1}=(-1)^m\ \Rightarrow\ I=2\pi i(-1)^m \end{gather} P.S. We have some videos with explanations about contour integrals. Check out our page: https://www.youtube.com/channel/UCcOPzqeuB8GTt1zfh6JKbSA
stokes-line
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Nice videos! That being said, self-promo is kinda a no-no here :( – Rushabh Mehta Mar 15 '20 at 00:09
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What are the assumptions on $m$ and $n$ though? If you're allowing only negative integers, then $m + n$ cannot be $-1$. If you're allowing any integers, then it's not true that the residue at infinity is non-zero only for $m + n = -1$. – Maxim Mar 15 '20 at 03:08
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Indeed, my blunder. My answer is only true for $m,\ n\leq 0$. If $m$ and $n$ are strictly negative then the answer is zero, of course. – stokes-line Mar 15 '20 at 05:39
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You said $n,m\lt0$. By Cauchy's integral formula, I get $2\pi i (1+1)=4\pi i$, when $n=m=-1$.
For other negative values of the exponents, it appears to be more complicated, and I don't see how to do it with CIF. Perhaps you could use the residue theorem.