Does $f(z)=\frac{e^z+\sin(z)}{z}$ have a complex antiderivative on $\mathbb{C}/\{0\}$

Question

The way the question was asked in our complex analysis course made me think it doesn't have. I wanna use Cauchy' Theorem to show that $\int_{\gamma}f(z)dz\neq0$ but I have a hard time finding a good path. I tried $\gamma_1(t)=e^{it}$ and $\gamma_2(t)=e^{-it}$ but it didn't seem to work.

Answer 1

Your idea is the right one. If $\int_\gamma f=0$ for all contours $\gamma$ that avoid the origin, you would indeed have a complex antiderivative.

Choosing the unit circle is just fine. Let $\gamma:[0,1]\to\Bbb C$ take $t\mapsto\exp(2\pi it)$.

Hint: you want to use the residue theorem to evaluate $\oint_{\gamma}\frac{\exp(z)+\sin(z)}{z}\,\mathrm{d}z$. What type of pole does it have at zero?