Evaluate coefficients of complex power series Evaluate the coefficients of this power series $$\frac{e^{tz}}{1-z}= \sum_{n=0}^{\infty} c_n (t) z^n$$ 
I rewrote it as $e^{tz} = (1-z) \sum_{n=0}^{\infty} c_n  z^n = \sum_{n=0}^{\infty} (c_n - c_{n-1}) z^n$, with some manipulation of the index of summation. 
By comparison to the power series of exponential  I find the recursion $$c_n - c_{n-1} = \frac{t^n}{n!}$$
How to find now the value of $c_1$ and $c_0$ (or $c_0$ and $c_{-1}$)?
 A: By plugging $z=0$ into $f(z)$ we can find $c_0$ and so
$$c_0=\frac{e^{t\cdot0}}{1-0}=1$$
$$c_n=c_{n-1}+\frac{t^n}{n!}$$
$$\therefore c_n=\sum_{k=0}^n \frac{t^k}{k!}$$
A: Your calculation 
$$
e^{tz} = (1-z) \sum_{n=0}^{\infty} c_n  z^n = \sum_{n=0}^{\infty} (c_n - c_{n-1}) z^n
$$
is only valid if you define $c_{-1} = 0$. Then
$$
\begin{align}
 c_{-1}  &= 0 \\
 c_n - c_{n-1} &= \frac{t^n}{n!} \text{ for } n \ge 0 \, .
\end{align}
$$
determines all $c_n$ uniquely.
If you want to avoid negative indices then you have to do the index manipulations carefully:
$$
e^{tz} = (1-z) \sum_{n=0}^{\infty} c_n  z^n = \sum_{n=0}^{\infty} c_n  - \sum_{n=0}^{\infty} c_n z^{n+1} =
 \sum_{n=0}^{\infty} c_n  - \sum_{n=1}^{\infty} c_{n-1} z^{n}
= c_0 + \sum_{n=1}^{\infty} (c_n - c_{n-1}) z^n
$$
And now the comparison of the coefficients of the power series gives
$$
\begin{align}
 c_0 &= 1 \\
 c_n - c_{n-1} &= \frac{t^n}{n!} \text{ for } n \ge 1 \, .
\end{align}
$$
which again determines all $c_n$ uniquely.
A: With Cauchy - product we have
$$\frac{e^{tz}}{1-z}= (\sum_{n=0}^{\infty} \frac{t^n}{n!}z^n)(\sum_{n=0}^{\infty}z^n)=\sum_{n=0}^{\infty}\phi(n,t,z)$$
with $\phi(n,t,z)=\sum_{k=0}^n\frac{t^k}{k!}z^k \cdot z^{n-k}=(\sum_{k=0}^n\frac{t^k}{k!})z^n.$
