Find all analytic functions on $\mathbb{D}$ s.t $f(\frac{1}{n})+f''(\frac{1}{n})=0$.
My attempt: Using uniqueness theorem, I'm looking for all analytic functions satisfying $f=-f''$. This implies that $f$ satisfies:$$f=f^{(4n)}=-f^{(4n+2)},f'=f^{(4n+1)}=-f^{(4n+3)}$$Expanding $f$ at $0$ gives: $$f=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^{n}=\sum_{4n}\frac{f^{(4n)}}{(4n)!}z^{4n}+\sum_{4n+1}\frac{f^{(4n+1)}}{(4n+1)!}z^{4n+1}+\sum_{4n+2}\frac{f^{(4n+2)}}{(4n+2)!}z^{4n+2}+\sum_{4n+3}\frac{f^{(4n+3)}}{(4n+3)!}z^{4n+3}$$
And rewriting gives: $$f=\sum_{4n}\frac{f(0)}{(4n)!}z^{4n}+\sum_{4n+1}\frac{f'(0)}{(4n+1)!}z^{4n+1}-\sum_{4n+2}\frac{f(0)}{(4n+2)!}z^{4n+2}-\sum_{4n+3}\frac{f'(0)}{(4n+3)!}z^{4n+3}$$
I wasn't able to get a clear description of all functions satisfying this. I do suspect that the functions $e^{iz},e^{-iz}$ span all functions satisfying these conditions, but I wasn't able to prove it. Any hint would be appreciated.