Antiderivative of complex function on unit circle

Question

I need advice with the function :

$$f(z) = \frac{e^z-1}{\sin z}$$ on defined on the unit circle, and seeing if it has a holomorphic antiderivative on the unit circle, so $F'(z)=f(z)$

I know this is equivalent to $\int_\gamma f(z)dz=0$ on all closed paths $\gamma$. Can this be done without residue theorem?

Answer 1

Let $1 <r<\pi$ and define $f(0)$ to be $1$. Then $f$ is analytic in $\{z: |z|<r\}$ and it has no zeros there. Hence, it has an anti-derivative throughout $\{z: |z|<r\}$.