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Suppose we have the following contour integral, in the complex plane: $$ \int_{\gamma} \frac{e^{\frac{1}{z}}}{z^{2}} \; dz $$ where $\gamma (t) = e^{it}$ for $0 \leq t \leq 2 \pi$.

To solve this, I would express the integral in the form $$ \int_{\gamma} \frac{f(z)}{z-z_{0}} \; dz $$ Where we would have $$ f(z) = \frac{e^{\frac{1}{z}}}{z} \\ z_{0} = 0 $$

Then, by Cauchy's integral formula, we have that $$ \int_{\gamma} \frac{f(z)}{z-z_{0}} \; dz = 2 \pi i \cdot n( \gamma , z_{0} ) \cdot f(z_{0}) $$ However, $f(z_{0})$ is undefined for $z_{0} = 0$. What can I do to avoid this problem?

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You can use residue theorem.

$$e^{1/z}=1+\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}+\dots$$ $$\frac{e^{1/z}}{z^2}=\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{2z^4}+\frac{1}{6z^5}+\dots$$

So the residue of $\frac{e^{1/z}}{z^2}$ in $z=0$ is $0$. So $\int_{\gamma} \frac{e^{\frac{1}{z}}}{z^{2}} \; dz =0$.

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