Prove simple closed curves $f$'s exist, so $\Gamma = C-\sum_{i=1}^{k}{f_i}$ satisfies $ \int_{\Gamma}{\frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz}=0$

Question

Let $C$ be the circle $C(0,2)$ traversed one time counter-clockwise. Prove that there exist $k\in \mathbb {Z}_+$ and $C^1$ simple closed cuves $f_1, \dots ,f_k$ such that the cycle $\Gamma = C-\sum_{i=1}^{k}{f_i}$ satisfies $$ \int_{\Gamma}{\frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz}=0$$

Note: $C^1$ means the curve has continuous derivatives for each $t$ within the curve's interval $[a,b]$.

What would be the simplest way to prove this?

Answer 1

The simplest way to prove it is by providing an example.

Note that the singularities of the integrand are at $a_{1,2}=\pm i,a_{3,4}=(-1\pm i\sqrt3)/2,a_5=0$.

Thus, let $f_n=C(a_n,\epsilon)$ where $1\le n\le 5$ and $\epsilon$ being sufficiently small.

Then, clearly, $C-\sum f_n$ encloses no singularities. Therefore, by Cauchy’s integral theorem, $$\oint_{\Gamma} \frac{z^3e^{1/z}}{(z^2 + z + 1)(z^2 + 1)}dz =0$$