Can someone help me? I started but i have problem

Question

$\int_Φ \frac{e^{2z} } {z^4(z-2)}dz$ where Φ:[0,2Pi] and $Φ(t)=3e^{it} $ so we have $z=3e^{it}$ and $dz=3ie^{it}dt$ I want to calculate $ \int_0^{2Pi} \frac{e^{6e^{it}}} {e^{4}(3e^{it}-2)} 3ie^{it}dt = 3i \int_0^{2Pi} \frac{e^{7e^{it}}} {e^{4}(3e^{it}-2)} dt=3i \int_0^{2Pi} \frac{e^{7e^{it}}} {3e^{4it}-2e^{4}} dt= 3ie^{-4} \int_0^{2Pi} \frac{e^{7e^{it}}e^{-it}} {3-2e^{-it}} dt$ and now i take $e^{-it} =w$ and $-ie^{-it} dt=dw$ so we have $\frac{-3i} {ie^4}\int_0^{2Pi} \frac{e^{7w^{-1}}dw} {3-2w} dw$,and I dont know what should I do now

Answer 1

Hint: You have two poles, at $0$ and $2$ inside the circle of radius $3$, compute the residu and apply the residus formula.

Answer 2

You can use the fact that$$\frac{e^{2z}}{z^4(z-2)}=\frac{e^{2z}}{16(z-2)}-\frac{e^{2z}}{2z^4}-\frac{e^{2z}}{4z^3}-\frac{e^{2z}}{8z^2}-\frac{e^{2z}}{16z}$$and apply Cauchy's integral formula.