Let $f$ have a non-removable singularity at 0, and let $f$ be holomorphic on the punctured disk $B'(0,1)$.
I want to compute Ind$(\exp \circ f \circ \alpha, 0)$ where $\alpha(t) = 1/2 e^{it}$, for $0 \leq t \leq 2 \pi$.
So I tried just using the definition
Let $\beta(t)= \exp \circ f \circ \alpha$. Since all the functions here are holomorphic on the domain $\beta$ is a piecewise smooth curve. Since $\beta(0)=\beta(2\pi)$ the curve $\beta$ is closed.
\begin{equation}\begin{split}\text{Ind}(\beta,0) = \frac{1}{2 \pi i} \int_\beta \frac{dz}{z-0} &= \frac{1}{2 \pi i} \int_{f \circ \alpha} \frac{\exp(\zeta)'}{\exp{(\zeta})}d\zeta \\ &= \frac{1}{2 \pi i} \int_{f \circ \alpha} \frac{\exp(f(\alpha))(f(\alpha))'}{f(\alpha)}d\zeta \\ &= \frac{1}{2 \pi i} \int_{\alpha(0)}^{\alpha(2\pi)} \frac{\exp(f(\alpha))f'(\alpha)\alpha'(t)}{f(\alpha)}dt \end{split}\end{equation}
I've no idea what $f$s values could be so this doesn't look that helpful. Since there's only one singularity, I worked out the Residue theorem too. I used the identity function over the curve $\beta$ \begin{equation}\begin{split}\frac{1}{2 \pi i} \int_\beta z dz &= \text{Ind}(\exp(f(\alpha)),0)\text{Res(e,0)} \\ \frac{1}{2 \pi i} z^2 \Big|_{\beta(0)}^{\beta(2\pi)} &= \text{Ind}(\exp(f(\alpha(t))),0)\frac{1}{2 \pi i} \int_{|z|=r} e^z dz. \end{split}\end{equation}
The RHS has an integral I can probably look up in the notes or the internet (is it just $2\pi$?) The LHS should be $0$ since $\beta(0)=\beta(2 \pi)$. So either I made a mistake or this isn't helpful.
The only other thing I tried was using a Laurent series for $f$. That still leaves me in the position of having to integrate the unknown function $f$ over a curve.
The question has a second part too, Show that $\exp \circ f$ does not have a pole at $0$. But I don't think that's related. Also, FYI non-removable singularities of a function are essential singularities of the composition function