Taylor series of $e^{i \sin z}$ How can I find the Taylor series of at $z=0$ (where $z$ complex ) of:
$e^{i \sin z}$?
What I wrote is:
$$e^{i \sin z}= \sum_0^\infty \frac{(i \sin z)^n}{n!}$$
Is that right? And how it can be more simplified?
 A: You need to exponentiate the taylor series of $\sin z$
$$\sum_i \frac{i^n}{n!}\bigg(\sum_j (-1)^j \frac{z^{2j+1}}{(2j+1)!}\bigg)^n$$
Exponentiating a convergent series is equivalent to autoconvolving the coefficients.
But it will be too difficult I think. For the first few terms you can expand the series and exponentiate yourself. You will not be able to use all the terms.
I actually think finding the derivatives evaluated at z=0 would be the easiest way. I would use this:

Software gives this result:

A: Here is an approach to get a closed form.
One  may observe that
$$
e^{e^x-1}=\sum_0^{\infty}\frac{B_n}{n!}x^n
$$ where $B_0=1, B_1=1, B_2=2, B_3=5, \ldots, B_n$ are the Bell numbers given by
$$
B_n=\sum_{k=1}^n\sum_{j=0}^k\frac{(-1)^{k-j}}{k!}\binom{k}{j}j^n.
$$
Then one may write
$$
\begin{align}
e^{i \sin z}&=e^{\large e^{ (iz-\ln 2)}}\times e^{\large e^{- (iz+\ln 2+i\pi)}}\\\\
&=e^2\sum_0^{\infty}\frac{B_n}{n!}(iz-\ln 2)^n \times \sum_0^{\infty}(-1)^n\frac{B_n}{n!}(iz+\ln 2+i\pi)^n\\\\
&=e^2\sum_{n,k\geq 0}(-1)^k\frac{B_n}{n!}\frac{B_k}{k!}(iz-\ln 2)^n \times (iz+\ln 2+i\pi)^k\\\\
&=e^2\sum_{n,k,\ell_0,\ell\geq 0}(-1)^kB_nB_k\frac{(-\ln 2)^{n-\ell_0} \times (\ln 2+i\pi)^{k-\ell+\ell_0}}{\ell_0!(n-\ell_0)!(k-\ell+\ell_0)!(\ell-\ell_0)!}(iz)^{\ell}.
\end{align}
$$
