# Cauchy's formula: $\int_\gamma \frac{e^{z}}{z(z-3)} dz$

trying to compute the integral $\int_\gamma \frac{e^{z}}{z(z-3)} dz$, where $\gamma:[0,2\pi]\to\mathbb{C}, \gamma(\theta)=2e^{i\theta}$ but not sure where to begin. I know, from Cauchy's formula, that $f(b) = \frac{1}{2{\pi}i} \int_\gamma \frac{f(a)}{a-b}da$, so I attempted to proceed by letting $f(a) = e^{a}$ and realizing the denominator consists of product terms $z$ and $z-3$, both of the form $a-b$, with $a=z$ and $b=0$ and $b=3$ in each term. However, I'm not sure how to deal with the two product terms, nor am I sure in my $f(a) = e^{a}$ substitution. Any help is greatly appreciated!

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## 2 Answers

Use $f(z) = \frac{e^z}{z-3}$, and $b = 0$. Then you have the right integrand, and can use the theorem.

I learned this basic application from the example on the wikipedia article on the theorem. Depending on the book you use, wikipedia might be better for the concrete understanding (it was for me).

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Why does $\gamma$ contain $3$? – Dennis Gulko Oct 9 '12 at 21:23
Sorry, it doesn't. For some reason, I thought the radius was $2e$. – Arthur Oct 9 '12 at 21:26

Observe that $|3|>2$, hence $g(z)=\frac{e^z}{z-3}$ is analytic in $|z|<3$. So you can rewrite you integral in the form familiar from Cauchy's theorem: (applying the theorem to $g(z)$) $$\int_\gamma\frac{g(z)}{z-0}dz=2\pi ig(0)=-\frac{1}{3}$$

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