Let $f$ be holomorphic function. Let $f''(z) + f(z) = 0$ for all $z \in \Omega$ and $f'(c) = f (c) = 0$ for some $c \in \Omega$ . Show that $f=0$ Problem
Let $f$ be holomorphic in a region $\Omega$. Suppose that $f''(z) + f(z) = 0$ for all $z \in \Omega$
and $f'(c) = f(c) = 0$ for some $c \in \Omega$
. Show that $f=0$ identically.
It must be an application of Identity theorem but I am unable to figure out how to use the theorem!
Any hint will be appreciated.
 A: We need that $\Omega$ is connected. In this case, it would be enough to show that the Taylor expansion of $f$ at $c$, say $f(z)=\sum_{n=0}^{+\infty} a_n(z-c)^n$, is such that $\forall n \in \mathbb{N}, a_n =0$. By $f''+f=0$ we get

*

*$\forall n \in \mathbb{N}, a_{2n}=a_0\frac{(-1)^n }{(2n)!}$.

*$\forall n \in \mathbb{N}, a_{2n+1}=a_1\frac{(-1)^n }{(2n+1)!}$.

The conclusion follows since $f(c) = 0$ implies $a_0 = 0$ and $f'(c)=0$ implies $a_1 = 0$.
A: Let $z \in \mathbb C$ and $\epsilon >0$ such that $c + (-\epsilon, \epsilon)z \subset \Omega$. Then consider $g:t\in (-\epsilon, \epsilon) \mapsto f(c+tz)$. This is a smooth map and we have :
$$\forall t\in (-\epsilon,\epsilon), g''(t) + c^2 g(t) = 0$$
Furthermore, $g'(0)=  g(0) = 0$. Therefore $g$ is identically $0$.
Since $\Omega$ is locally convex, we can show that $f$ is zero on a neighborhood of $c$. Since $f$ is holomorphic and $\Omega$ is connected, we know that $f =0$
A: An alternative solution: Consider the function $g = f'-if$ in $\Omega$. Then
$$
 \frac{d}{dz} e^{iz} g(z) = e^{iz} (g'(z)+ig(z)) = e^{iz}( f''(z)+f(z)) = 0
$$
which implies (since $\Omega$ is connected) that $e^{iz} g(z)$ is constant in $\Omega$. But $g(c) = 0$, therefore $g$ is identically zero in $\Omega$.
Now we repeat the argument:
$$
\frac{d}{dz} e^{-iz} f(z) = e^{-iz} (f'(z) -if(z)) =  e^{-iz}  g(z) = 0
$$
which implies that $e^{-iz} f(z)$ is constant. But $f(c) = 0$, therefore $f$ is identically zero in $\Omega$.
Remark: The idea of this approach is to “solve”
$$
 0 = f'' + f = \left( \frac{d^2}{dz^2} + 1\right) f(z)
= \left( \frac{d}{dz} + i\right) \left( \frac{d}{dz} - i\right)f(z)
$$
for $f$.
