Using C-G Theorem or CIF theorem evaluate this integral

Question

There are multiple singularities in this region, do I just split this into a partial fraction and then evaluate using CIF? $$I =\oint_{|z-3|=2} \frac{\exp(3z)\,dz}{z(z-2)(z-4)^2}.$$

Answer 1

Hint:

Among denominators roots $z=0$, $z=2$ and $z=4$ only $z=2$ and $z=4$ are in circle $|z-3|=2$, so with residue theorem $$\oint_{|z-3|=2} \frac{e^{3z}}{z(z-2)(z-4)^2}\text{dz}=2\pi i\left(\text{Res}(f,2)+\text{Res}(f,4)\right)$$ and $$\text{Res}(f,2)=(z-2)\frac{e^{3z}}{z(z-2)(z-4)^2}\Big|_{z=2}=\dfrac{e^6}{8}$$ Can you take it from here!

Answer 2

Yes, your approach is correct (but this evaluation can be in a more straightforward way).

The partial fraction decomposition gives $$f(z)=\frac{1}{z(z-2)(z-4)^2}=\frac{A}{z}+\frac{B}{z-2}+\frac{C}{z-4}+\frac{D}{(z-4)^2}.$$ Since the poles inside $|z-3|=2$ are $2$ and $4$, then by the residue theorem $$I=\oint_{|z-3|=2} f(z)\exp(3z)dz\\=2\pi i \left( B\,\mbox{Res}\left(\frac{e^z}{z-2},2\right) +C\,\mbox{Res}\left(\frac{e^z}{z-4},4\right) +D\,\mbox{Res}\left(\frac{e^z}{(z-4)^2},4\right) \right).$$ After the decomposition, the residues are easy to evaluate because for $z_0\in\mathbb{C}$ and $m\in\mathbb{N}^+$, $$\mbox{Res}\left(\frac{e^z}{(z-z_0)^m},z_0\right)= e^{z_0}\mbox{Res}\left(\frac{e^{z-z_0}}{(z-z_0)^m},z_0\right)=e^{z_0}$$ (see also the Cauchy's integral formula).