Taylor series for $e^z\sin(z)$ How can I write the Taylor series for $e^z\sin(z)$ at $z=0$ without making the procedure too complicated?
Isn't there an easier way than to compute it's derivatives and find a pattern?
 A: Yes, there are simpler methods based on the power series expansion of $\exp$. I indicate two equivalent methods, the first one being more elementary than the second one.
Method 1: Use Euler's formula $\sin(z) = \dfrac{e^{iz} - e^{-iz}}{2i}$.
Method 2: For all $x \in \Bbb R$,
$$
e^{x}\sin(x) = \Im(e^{(1+i)x}) = \sum_{n=0}^\infty\frac{\Im((1+i)^n)}{n!}x^n= \sum_{n=0}^\infty \frac{2^{n/2}\sin(\frac{n\pi}{4})}{n!}x^n.
$$
By analytic continuation, the identity still holds for $z \in \Bbb C$.
A: to find the $nth$ derivative of 
$$e^{az}\sin (bz+c)$$
differentiating this expression once
$$D^1=e^{az}(a\sin (bz+c)+b\cos (bz+c))$$
let $a=r\cos \theta$ and $b=r\sin \theta$. you get
$$D^1= e^{az}r\sin (bz+c+\theta)$$
if you observe you see
$$D^n= e^{az}r^n\sin (bz+c+n\theta)$$
where $r=\sqrt{a^2+b^2}$  and $\theta=\arctan(\frac{b}{a})$
now substitute for $a,b,c$.
A: I think the easiest procedure is the multiplication of power series. Let
\begin{align}
g(z)= & a_0+b_1z+a_2z^2+a_3z^3+\ldots+a_nz^n+\ldots\\ 
f(z)= & b_0+b_1z+b_2z^2+b_3z^3+\ldots+b_nz^n+\ldots
\end{align}
Then
\begin{align}
g(z)f(z)
=
&
(a_0b_0)
\\
+
&
(a_1b_0+a_0b_1)z
\\
+
&
(a_0b_2+a_1b_1+a_2b_0)z^2
\\
+
&
(a_0b_3+a_1b_2+a_2b_1+a_3b_0)z^3
\\
+
&
(a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0)z^4
\\
\vdots
&
\\
\vdots
&
\\
+
&
\sum_{i+j=k}a_ib_jz^k
\end{align}
implies
$$
g(z)f(z)=\sum_{k=1}^\infty\color{red}{\sum_{i+j=k}a_ib_j}z^k.
$$
In your case,
\begin{align}
e^z\sin(z)
=
&
\left(\sum_{\alpha=0}^{\infty}\frac{1}{\alpha!}z^\alpha \right)\left(\sum_{\beta=0}^{\infty}\frac{(-1)^{\beta}}{(2\beta+1)!}z^{2\beta+1} \right)
\\
=
&
\sum_{k=0}^{\infty}\;\;
\color{red}{\sum_{\alpha+(2\beta+1)=k}
\left[\frac{1}{\alpha!}\cdot\frac{(-1)^{\beta}}{(2\beta+1)!} \right]}
z^{k}
\end{align}
