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$$\int_{\gamma=(i,1)} \frac{z^3}{(z-i)^n} dz$$ for any $n\in\mathbb{N}$.

Can someone please help me answer this question as I cannot seem to get the right answer!

Please note that the Cauchy integral formula must be used in order to solve it.

Many thanks in advance!

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  • $\begingroup$ What is your contour? It doesn't make sense currently. $\endgroup$ – anon271828 Mar 15 '13 at 0:24
  • $\begingroup$ well that is exactly how the question appears in the book. $\endgroup$ – camilla Mar 15 '13 at 0:27
  • $\begingroup$ What book are you using? $\endgroup$ – anon271828 Mar 15 '13 at 0:27
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    $\begingroup$ Maybe it means a circle of radius 1 centered at $i$? $\endgroup$ – Chris Brooks Mar 15 '13 at 0:29
  • $\begingroup$ yes it means |z| = 1 and centred at i. i dont know what a contour is, thats why im asking for help im massively stuck!! $\endgroup$ – camilla Mar 15 '13 at 0:30
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Use the residue theorem. Since $\cfrac{z^3}{(z-i)^n}$ has a pole of order $n$ at $z=i$ and analytic everywhere other than $z=i$ in the domain $|z-i|<1$, by residue theorem, we have $$\int_{\gamma}\frac{z^3}{(z-i)^n}dz=2\pi ig(i),\text{where }g(z)=\frac{1}{(n-1)!}(z^3)^{(n-1)}$$

The residue theorem is obtained from Cauchy Integral formula.

By Cauchy Integral formula, we have $$2\pi if(z)=\int_C\frac{f(\zeta)}{\zeta-z}d\zeta$$ Differentiate both sides with respect to $z$ $n-1$ times, we get $$2\pi if^{(n-1)}(z)=(n-1)!\int_C\frac{f(\zeta)}{(\zeta-z)^n}d\zeta$$

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  • $\begingroup$ I can't use the residue theorem is has to be the Cauchy Integral Formula!! Sorry but thank you :) $\endgroup$ – camilla Mar 15 '13 at 0:45
  • $\begingroup$ @camilla Residue theorem and Cauchy Integral Formula are actually the same thing. Or, just attach the "two-line proof" to your homework $\endgroup$ – NECing Mar 15 '13 at 0:48
  • $\begingroup$ is there not a way of putting it into partial fractions to solve? this is what im used to - not understanding very well sorry!! btw thank you for your answers. $\endgroup$ – camilla Mar 15 '13 at 0:49
  • $\begingroup$ Contour integral is different from the real one. You can not directly take the anti-derivative. $\endgroup$ – NECing Mar 15 '13 at 0:51
  • $\begingroup$ okay, im a bit confused about this one, but would the answer be (2pi)i? $\endgroup$ – camilla Mar 15 '13 at 0:58
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Hint: Parametrize the contour as $z=i+e^{i\theta}$ for $0\leq\theta\leq 2\pi$. Then, by definition, you have $$\int_0^{2\pi}\frac{(e^{i\theta})^3}{((i+e^{i\theta})-i)^n}\cdot ie^{i\theta}\,d\theta=i\int_0^{2\pi}\frac{(e^{i\theta})^3}{e^{in\theta}}\cdot e^{i\theta}\,d\theta=i\int_0^{2\pi}(e^{i3\theta})e^{i(1-n)\theta}\,d\theta.$$Keep simplifying and you can solve it.

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  • $\begingroup$ is that using the cauchy integral forumula though?? Many thanks $\endgroup$ – camilla Mar 15 '13 at 0:37
  • $\begingroup$ should i not put it into partial fractions and use the cauchy formula twice? I have a different looking answer here! $\endgroup$ – camilla Mar 15 '13 at 0:42

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