TITLE: If $f$ is holomorphic and $\,f'' = f$, then $f(z) = A \cosh z + B \sinh z$
QUESTION [6 upvotes]: Suppose $f$ is holomorphic in a disk centered at the origin and $f$ satisfies the differential equation $$f'' = f.$$ Show that $f$ is of the form $$f(z)=A \sinh z + B \cosh z,$$ for suitable constants $A,B\in \mathbb C$.

Of course if $f$ were a function of one real variable, usual techniques to solve linear homogeneous differential equations (characteristic equations and so on) would quickly solve the problem. But using the same train of thought here, $f$ is a function of two real variables ($z = x + iy$), and so the same techniques cannot be applied directly. This is really a complex analysis problem.

REPLY [5 votes]: Let : $$g=f-f(0)\cosh-f'(0)\sinh$$
This function is holomorphic in the same disc $D=D(0,r)$ as $f$. 
You clearly have $g(0)=g'(0)=0$ and $g''=g$, so by induction $g^{(k)}(0)=0$ for all $k\geqslant 0$, and consequently
$$g(z)=\sum_{k\geqslant 0} \frac{g^{(k)}(0)}{k!}z^k=0$$
for all $z$ in a disc $D(0,r')$ with $0<r'\leqslant r$.
Hence $g=0$ on the domain $D$, i.e. $f=f(0)\cosh+f'(0)\sinh$ on $D$.