Evaluate $\int_{C}\frac{e^{3z}}{(z+1)^5}dz$

Question

$$\int_{C}\frac{e^{3z}}{(z+2)^5}dz\,\,\,\,\,\,\,\,C:|z|=3$$

My try:

Applaying Cauchy's formula

$$\int_{C}\frac{e^{3z}}{(z+2)^5}dz=2\pi ie^{-2\cdot 3}=\frac{2\pi i}{e^{6}}$$

I'm not sure in my attempt

Answer 1

Notice, your final answer is wrong.

$z=-2$ is a pole of fifth order inside the circle: $|z|=3$ & $f(z)=e^{3z}$

Cauchy's formula for $n$th order pole is given as $$\color{blue}{\int_C\frac{f(z)}{(z-a)^n}\ dz=\frac{2\pi i}{(n-1)!}\left[\frac{d^{n-1}}{dz^{n-1}}(f(z))\right]_{z=a}}$$

hence using Cauchy's formula, one should get $$\int_C\frac{e^{3z}}{(z+2)^5} \ dz$$$$=\frac{2\pi i}{4!}\left[\frac{d^4}{dz^4}(e^{3z})\right]_{z=-2}$$ $$=\frac{2\pi i}{4!}\left[3^4e^{3z}\right]_{z=-2}$$ $$=\frac{27\pi i}{4}e^{-6}=\color{red}{\frac{27\pi i}{4e^6}}$$