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I want to compute $$\int_{\gamma} \frac{e^z}{z^4-1} \, dz $$ where $\gamma$ is the unit circle centered at $1$. I'm asked to do this using Cauchy's integral formula. However, I need the denominator to be in the form $(z-a)^{n+1}$, how can I rewrite this function so that I can apply the formula?

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    $\begingroup$ I'm afraid you will have to extend your circle a little since all four singularities fall on that. $\endgroup$ – Vim Nov 30 '16 at 23:56
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    $\begingroup$ $z^4-1=(z-1)(z+1)(z-i)(z+i)$ $\endgroup$ – MattG88 Nov 30 '16 at 23:57
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You can factor the denominator using a difference of squares twice $$z^4-1=(z^2+1)(z^2-1)=(z+i)(z-i)(z+1)(z-1)$$ The only singular point in the interior of $\gamma$ is $z=1$ (this is easy to see if you draw a picture) and so $$\int_\gamma\frac{e^z}{z^4-1}=\int_\gamma\frac{e^z}{(z+i)(z-i)(z+1)(z-1)}dz=2\pi i \frac{e}{(1+i)(1-i)(1+1)}= \frac{ie\pi}{2}$$

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Define $$f(z) = \frac{e^z}{(z+1)(z^2+1)}$$ on the interior of your disk. Then, by Cauchy integral formula, $$\frac{1}{2 \pi i}\int_{\gamma} \frac{f(z)}{z-1} \mathrm{d}z = f(1) = \frac{e}{4}$$ so that your integral equals $ \pi e i/2$.

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  • $\begingroup$ You missed a factor of 2: $f(1)=e/4$ $\endgroup$ – user275377 Dec 1 '16 at 0:15

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