The Cauchy formula can be used to find the $n$th derivative of an analytical function $f(z)$ as $$ \frac{d^nf}{dz^n}\bigg|_{z=z_0}=\frac{n!}{2\pi i }\oint \frac{f(z)}{(z-z_0)^{n+1}}dz $$ which happens to be the coefficients of the Laurent series of $f(z)$.
I have a very similar formula that is given by the following contour integral
$$ \frac{n!}{2\pi i }\oint \frac{f(z^a)}{(z-z_0)^{n+1}}dz $$ However, note the power $z^a$ in the argument of $f(z^a)$. How can I write the left hand side of this function in terms of a derivative? i.e. can we still use the Cauchy formula?
If you let $g(z) = f(z^a)$, you can directly apply the Cauchy formula to $g$ and get : $$\left.\frac{d^ng}{dz^n}\right|_{z=z_0} = \frac{n!}{2\pi}\oint \frac{f(z^a)}{(z-z_0)^{n+1}} dz $$