Complex Ordinary Differential Equations I would like to understand a bit on complex ordinary differential equations, since I just learned some theorems on complex integration. So I proposed my-self to solve the following one:
$$f(z) = f'(z), \,\,\,\, f(1)=z_0$$
Let's suppose $f$ is analytic on $D$ such that the initial value is in. If it were a real ODE, that would could be solved through separation of variables. But that's not the case here, since we would have to integrate over a path.
Any ideas or even the solution would be appreciated. However, I also would like to read any book on this subject.
Thanks
EDIT
Well, as far as I know. There may be two ways: 1) Integrate $f$ over a rectifiable path and use the FTC for complex functions or 2) Separate $f$ into real and imaginary parts. The thing is, how do to the first way. Any ideas or comments? 
 A: Here's a way of seeing an approach which doesn't just reduce to the real case, and uses some salient elementary features of complex functions. To make our computation straightforward later and because the normalization is at $1$, let's assume that $f$ satisfies this ODE and is defined on an open set containing the closed unit disk. If $f$ satisfies this ODE, as it's holomorphic we're justified in taking a Taylor series expansion on some ball containing the closed unit disk at $0$, $$f(z) = \sum_{n=0}^\infty c_n z^n.$$ By elementary complex analysis, we're free to differentiate term-by-term and our ODE becomes $$\sum_{n=0}^\infty c_n z^n = \sum_{n=0}^\infty (n+1)c_{n+1}z^{n},$$ and so by linear independence of $z^n$ $(n = 0, 1, ...)$, we get that for all $n \geq 0$, $c_n = (n+1) c_{n+1}$. Induction shows this implies $c_n = c_0/n!$. So on an open ball of radius greater than $1$, $f(z) = c_0 \sum \frac{z^n}{n!} = c_0\exp(z).$ Normalization shows that then $f(z) = \frac{z_0}{e} \exp(z)$ on this ball, hence the entire connected component of the domain of $f$ containing $0$ by the identity principle.
A: Here is another point of view for this ODE, using the idea that the unique solution must look like an exponential.


*

*Assume that $f:D\rightarrow \mathbb C$ is a solution, with $D$ a connected set containing $1$ and let $g$ be the function given by $g(z)=f(z)e^{-z}$ on $D$.

*$g$ is holomorphic and for any $z\in D$, $$g'(z)=f'(z)e^{-z}-f(z)e^{-z}=0$$ since $f'=f$.


Now, one can easily deduce that $g$ is constant meaning that $f(z)=Ce^z$ with $C=z_0/e$ because $f(1)=z_0$.
