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Compute the contour integral $$∫_{|z|=1} \frac{e^{i(1+z)}}{z^{10}}dz$$

I am stuck at solving the integral. I know there is a singularity at z=0 and therefore we cannot apply the Cauchy Theorem directly. I used the parametrization $z=e^{it}$ and $dz=ie^{it}$ where $R=1$.

However my integration resulted into a mess and i cannot seem to reach the final answer.

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Have you learned the Residue Theorem? – Alan Guo Oct 21 '12 at 23:48
No i haven't learnt the residue theorem yet. We have learnt the generalized Cauchy theorem though – Khanak Oct 21 '12 at 23:50
up vote 1 down vote accepted

By Cauchy integral formula, we have that if $f(z)$ is holomorphic inside $\gamma$ and $a$ is inside $\gamma$, then $$f^{(n)}(a) = \dfrac{n!}{(2 \pi i)}\int_{\gamma} \dfrac{f(z)}{(z-a)^{n+1}} dz$$ In our case, $f(z) = \exp(i) \exp(iz)$, $n = 9$, $a = 0$ and $\gamma$ is the contour $\vert z \vert = 1$.

Also, recall that $$\dfrac{d^n (\exp(az))}{dz^n} = a^n \exp(az)$$ Now you should be able to get your answer.

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