Calculating $\oint_{|z| = 3} \frac{z^3e^{1/z}}{z^3+1}dz$ using a substitution

Question

So I've tried to calculate $$\oint_{|z| = 3} \frac{z^3e^{1/z}}{z^3+1}dz$$ using the substitution $w = \frac{1}{z}$ but I couldn't do anything after that. I believe I'm supposed to get to a point where I can apply the Residue Theorem to calculate the integral. The solution is $-2\pi i$ Any tips?

Answer 1

With the substitution $z\to \frac{1}{w}$ we have that $\frac{z^3 e^{1/z}}{z^3+1}\,dz$ is mapped into $$-\frac{e^w}{w^5+w^2}\,dw $$ and the integration range $|z|=3$, counter-clockwise oriented, is mapped into the integration range $|w|=\frac{1}{3}$, clockwise oriented. It follows that $$ \oint_{|z|=3}\frac{z^3 e^{1/z}}{z^3+1}\,dz = \oint_{|w|=\frac{1}{3}}\frac{e^w}{w^5+w^2}\,dw =2\pi i\cdot\text{Res}\left(\frac{e^w}{w^5+w^2},w=0\right)$$ where $w=0$ is a double pole for the new integrand function, and the only one enclosed by the integration path. Since in a neighbourhood of the origin we have $$ \frac{e^w}{1+w^3} = 1+w+O(w^2) $$ the wanted residue equals $1$ and $$ \oint_{|z|=3}\frac{z^3 e^{1/z}}{z^3+1}\,dz = \color{red}{2\pi i}.$$

Answer 2

If you know about Laurent series you may rewrite the integrand as $$ e^{1/z} \left(1+ \frac{1}{z^3}\right)^{-1}= (1+\frac{1}{z} + \frac{1}{2! z^2} + ...) (1-\frac{1}{z^3} +...) = 1+\frac{1}{z} + O(\frac{1}{z^2})$$ (the Laurent series converging for $|z|=3$) and the result follows by residue calculus.