Complex Integral $\frac{e^z}{z-1}$

Question

I'm looking to evaluate the following integral

$\int_{C(0,2)} \frac{e^z}{z-1} dz$.

So if I let $f(z)=\frac{e^z}{(z-1)^1}$ then:

$e^z$ is entire and the root of $z-1$ is $1$ and $1 \in D(0,2)$.

But $f(z)$ has a pole at $z=1$.

Does the Cauchy Integral theorem apply here. I haven't covered residues yet so that shouldn't be required. Any help would be appreciated.

Answer 1

The following is what I have since attempted.

$\int_{C(0,2)} \frac{f(z)dz}{z-z_0} = 2\pi i f(z_0)$

Hence, $f(z)=e^z$ and $z_0=1$

$\int_{C(0,2)} \frac{e^z}{z-1} dz= 2\pi i f(1)$

=$2\pi i e^1$

=$2\pi i e$