Analytic function $f,$ such that $f(0) = 1$ and $f'(z) = zf(z),$ for all $z \in \mathbb{C}$ I'm trying to find an example of an analytic function $f$ satisfying the IVP 
$$
f'(z) = z\,f(z), \quad
f(0) = 1, 
$$
and for all $z \in \mathbb{C}$, but I'm somewhat at a loss of the best way to proceed. 
Any hints or suggestions would be greatly appreciated.

Observe that
$$
f'(z)=zf(z)\,\,\Rightarrow\,\,\mathrm{e}^{-z^2/2}\big(f'(z)-zf(z)\big)=0
\,\,\Rightarrow\,\,\big(\mathrm{e}^{-z^2/2}f(z)\big)'=0,
$$
and thus $\mathrm{e}^{-z^2/2}f(z)$ is constant. In particular
$$
\mathrm{e}^{-z^2/2}f(z)=\mathrm{e}^{-0^2/2}f(0)=1,
$$
and thus
$$
f(z)=\mathrm{e}^{z^2/2}.
$$
