I'm having trouble evaluating the following contour integral:
$$I=\oint \frac{z^3e^{\frac{1}{z}}}{1+z^3}dz\ ;\ \ \ \ \ \ \text{for the anticlockwise circle defined by: }\quad |z|=3$$
What I've done is to realice that all singularities, including the essential one, fall within the circle:
My first step was to factorize de denominator like so:
$$1+z^3=\left(z + 1\right) \left(z-\frac{1+ \sqrt{3}\ i}{2}\right)\left(z-\frac{1-\sqrt{3}\ i}{2}\right)$$
since the numerator is well defined in all the solutions of the above polinomial, I can evaluate the residues without problem, evaluating the numerator where needed. This residues are $R_1, R_2$ and $R_3$
For the essential singularity this was what I did:
Since the problem is un the numerator, I wrote down a Laurent series for it…
$$z^3e^{\frac{1}{z}}=z^3\sum_{n=0}^{\infty}\frac{1}{n!z^n}=\sum_{n=0}^{\infty}\frac{z^{3-n}}{n!}$$
Here is the part that apparently is wrong:
I considered tha the residues in this point is the factor in the series for which the exponent equals -1, since the denominator of the integral is 1 in $z=0$, this residue which I will name $R_4$ is:
$$R_4=\frac{1}{4!}$$
Hence the integral is:
$$I=2\pi i(R_1+R_2+R_3+R_4)$$
Does anyone know if this is the correct solution?
Thank you all in advance.
Best Answer
Yes, $I=2\pi i(R_1+R_2+R_3+R_4)$, but $R_4$, your residue evaluation at $z=0$ is wrong. It should be the coefficient of $1/z$ in the product $$\frac{z^3}{1+z^3}\cdot e^{1/z} =\sum_{n=0}^{\infty}(-1)^n z^{3n}\cdot\sum_{n=0}^{\infty}\frac{1}{n!z^n}$$ that is $$\frac{1}{4!}-\frac{1}{7!}+\frac{1}{10!}-\frac{1}{13!}+\dots.$$
Since the function to integrate is holomorphic outside the circle $|z|=3$, it is easier to consider the residue at infinity, $$I=\oint_{|z|=3} \frac{z^3e^{\frac{1}{z}}}{1+z^3}dz=2\pi i\cdot\text{Res}\left(\frac{(1/z)^3e^{z}}{z^2(1+1/z^3)},0\right)=2\pi i\cdot\text{Res}\left(\frac{e^{z}}{z^2(z^3+1)},0\right)=2\pi i.$$