Evaluate the integral$$\int_{C}\frac{z^{3}e^{\frac{1}{z}}}{1+z^{3}}dz$$where C is the circle $\left|z\right|=3$, describe in the positive sense.
I want to clarify some points here. The singularities of the given function are $0,-1,e^{\frac{\pi i}{3}}$ and $e^{\frac{5\pi i}{3}}$. Hence, I have to find the residues on each of the singularities. I have problem in calculating the residue at $0$, I am not sure if Res$(f,0)=0$. For the rest of the singularities, I got no problem since I can factor the denominator into linear factors.
So, is Res$(f,0)=0$?
Best Answer
Since, near $0$,$$\frac{z^3}{1+z^3}=z^3-z^6+z^9-z^{12}+\cdots\tag1$$and$$e^{1/z}=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots,\tag2$$the residue at $0$ of$$\frac{z^3e^{1/z}}{1+z^3}\tag3$$is the coefficient of $\dfrac1z$ of the product between $(1)$ and $(2)$, which is$$\frac1{4!}-\frac1{7!}+\frac1{10!}-\cdots\tag4$$Besides, the sum of the residues of $(3)$ at $-1$, $e^{\pi i/3}$ and $e^{5\pi i/3}$ is equal to$$\frac{-1+e^{3/2}\left(\cos\left(\frac{\sqrt3}2\right)+\sqrt3\sin\left(\frac{\sqrt3}2\right)\right)}{3e}.\tag5$$So, your integral is $2\pi i$ times the sum of $(4)$ with $(5)$.