# 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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Use $f(z) = \frac{e^z}{z-3}$, and $b = 0$. Then you have the right integrand, and can use the theorem.
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 $g(z)$) $$\int_\gamma\frac{g(z)}{z-0}dz=2\pi ig(0)=-\frac{1}{3}$$