Find the Taylor series centered at $z_0$ of the function $f(z) = \sin(z^2)$. Find the Taylor series centered at $ z_0$ of the function $f(z) = \sin(z^2)$.
Solution:
$z$ is replaced by $z^2$ in the well-known expansion
$$ \sin z =
\sum_{n=0}^{ +\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)!}\qquad ( \forall z \in \mathbb C). $$ Thus,
$$ f(z) = \sin (z^2) = \sum_{n=0}^{ +\infty}\frac{(-1)^n(z^2)^{2n+1}}{(2n+1)!}=\sum_{n=0}^{ +\infty}\frac{(-1)^nz^{4n+2}}{(2n+1)!} \qquad( \forall z \in \mathbb C). $$ Is this correct?
I leave the exercise until there? If the procedure is wrong or if you have to continue developing, I hope you can help me.

The fact that $\sin(z)=(e^{iz}-e^{-iz})/2i$ allows considerable simplification, at the level of hand computation.
Namely, it suffices to know the power series of $e^{iz}$ at $z_o$, for which we have the Taylor-Maclaurin-Cauchy expansion
$$
e^{iz} \;=\; \sum_{n\ge 0} {d^n\over dz^n}e^{iz}\Big|_{z_o} {1\over n!} (z-z_o)^n
$$
and the derivative is easy to understand... :)
EDIT: depending on exactly what one wants, replacing $z$ by $z^2$ in a power series expansion may be "good". Or, sure, possibly replacing $z-z_o$ by $z^2-z_o$ and then doing the ugly algebra is what's demanded. My point is that this is feasible, by hand computation, but that the more details one wants, in awkward variants, the more unpleasant the computations become. My sincere reaction to such a question is "yes, I could compute those coefficients in finite time" (and maybe some comments about rationality)...
