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Evaluate $$\int_{C}{e^\frac{1}{z}dz}$$ where $C:|z|=1$ using the Residue Theorem

I found out that I don't need to use the Theorem at all.

Since $e^\frac{1}{z}$ is analytic over the entire complex region, thus also inside the unit circle which is simply connected, by the Morera Theorem I state that

$$\int_{C}{e^\frac{1}{z}dz}=0$$

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    $\begingroup$ I think there is a pole at $z=0$ $\endgroup$ Jan 10, 2020 at 23:36

1 Answer 1

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The function $e^{\frac{1}{z}}$ is not analytic. It has an essential singularity at $0$, which makes it normally difficult to deal with. However, we know that we have a power series representation for the exponential as

$$e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$$

We can then write

$$e^{\frac{1}{z}}=\sum_{n=0}^\infty\frac{1}{z^n}\frac{1}{n!}$$

The residue of this function is therefore the value we get when $n=1$, so

$$Res(e^{\frac{1}{z}})=1$$

Thus,

$$\oint_{|z|=1}e^{\frac{1}{z}}dx=2\pi i$$

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