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Evaluate: $$ \int_c {\sin z\over z^6}$$ Where, $c$ is a circle of radius $2, |z| = 2$. Don't understand why ${\pi i \over 5!} $ (Answer from book). $$ {\sin z \over z^6} = {1 \over z^6}\left [ z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040}+\frac{z^9}{362880}-\frac{z^{11}}{39916800}+O(z)^{12} \right ]$$

What happens to $ \int_c {1 \over z^5}$ and $- {1\over 3!} \int_c {1 \over z^3} $?

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You should use the Cauchy's differentiation formula –  Sasha Aug 27 '12 at 0:17
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... I don't understand why two terms vanishes. –  Monkey D. Luffy Aug 27 '12 at 0:19
    
Try calculating the integral $$\int_{|z|=2} \frac{1}{z^n} \,dz$$ directly by parameterizing the path (let $z = 2e^{i\theta}$ in the integral). You should get $0$ for every $n$ but $-1$. That's why those two terms vanish, and it's also the reason the residue is defined the way it is. –  Antonio Vargas Aug 27 '12 at 0:23
    
@AntonioVargas what about $$\int_{|z|=a} \frac{1}{(z-b)^n} \,dz$$ –  Monkey D. Luffy Aug 27 '12 at 0:33
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It depends on whether $|b| < |a|$ or $|b| > |a|$. In the first case the same thing happens--you only get a nonzero answer when $n=1$. In the second case you always get $0$. If you haven't learned them yet, these kinds of things will become more clear once you study Cauchy's theorem for integrating analytic functions on simple closed loops and then the residue theorem. Edit: Just realized I made a typo in my first comment. I definitely meant $n=1$, not $n=-1$. –  Antonio Vargas Aug 27 '12 at 0:38

2 Answers 2

up vote 2 down vote accepted

Hint: for a function $\,f(z)\,$ holomorphic within the domain inclosed by a closed smooth path $\,\gamma\,$, we have $$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz$$

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so they are zero by Cauchy differentiation formula? –  Monkey D. Luffy Aug 27 '12 at 0:21
    
$$ f^4 (0) = 0 = {4 ! \over 2 \pi i}{\int_c {1 \over z^5} dz}$$ –  Monkey D. Luffy Aug 27 '12 at 0:22
    
First, you have $\,f(z)=\sin z\,$ , second you have $\,n=5\,$ , not $\,4\,$ , thus:$$\oint_c\frac{\sin z}{z^6}dz=\frac{2\pi i}{5!}(\sin z)^{(5)}(0)$$ Now just remember that the derivatives of $\,\sin z\,$ are periodic modulo $\,4\,$ , so $\,(\sin z)^{(5)}=\sin'z=\cos z\,$ –  DonAntonio Aug 27 '12 at 1:45

Your series expansion is the same as $$ \frac{1}{z^5} - \frac{1}{6z^3} + \frac{1}{120z} - \frac{z}{5040} + \frac{z^3}{362880} -\cdots $$ If you integrate term-by-term, you get $0$ in every case except the third term.

(I'd have mentioned the identity that DonAntonio posted but for the fact that he already posted it, so I'm giving you a different point of view here. Learning that identity is virtually obligatory if you want to say you know this topic. And at least one way of proving it is very nice too: see the "Remarks" section of this article.)

Later edit: Your initial concern may have been why things like $$ \int_C \frac{dz}{z^n} $$ vanish when $n\in\{-2,-3,-4,\ldots\}$. If one has established that it doesn't matter what $C$ is as long as it winds once counterclockwise around the point where the function blows up, then we may as well take it for convenience to be the unit circle centered at $0$. Thus we have $$ \int_C \frac{dz}{z^n} = \int_0^{2\pi} \frac{i\theta e^{i\theta}\,d\theta}{e^{ni\theta}} = i\theta\int_0^{2\pi} \frac{d\theta}{e^{(n-1)i\theta}} = i\theta\int_0^{2\pi} e^{(1-n)i\theta}\,d\theta. $$ You can now apply the fundamental theorem of calculus, bearing in mind that $n-1>0$. If you have qualms about whether the F.T.C. works for complex-valued functions of a real variable, you can separate real and imaginary parts.

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thanks!! it was unknown to me –  Monkey D. Luffy Aug 27 '12 at 0:44

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