I'm trying to solve the following problem:
Given an analytic function $f: D \subset \mathbb{C} \to \mathbb{C}$, if $f^{(n)}(z) = 0$ for some $n \in \mathbb{N}$ and for all $z \in \mathbb{D}$, then $f$ is a polynomial of degree less than $n$.
I wanted to use induction to solve this problem. I managed to show that this is true for the base case of $f' = 0$. I then wanted to prove the inductive step holds.
To do this, I define the function $F = f'$, so that $f^{(n+1)} = F^{(n)}$ and I can apply the induction hypothesis on $F$. By doing this I get that $$ F = \sum_{k=0}^{n-1} a_kz^k = \sum_{k=0}^{n-1} \alpha_k r^k \cos(k\theta) + i \sum_{k=0}^{n-1} \alpha_k r^k + \sin(k\theta) \tag{1} $$ for some $a_k \in \mathbb{C}$ (possibly equal to $0$), $r = |z|$ and $\theta = \arg(z)$. From here I just need to recover $f$ to finish the problem, but here's where I ran into trouble.
The only way I know I can relate $f$ and $F$ is by using the Cauchy-Riemann equations, i.e., if $f = u(r, \theta) + iv(r, \theta)$, then \begin{align} F = f' &= \frac{\partial}{\partial r}\left(\left[\cos(\theta)u + \sin(\theta)v\right] + i \left[\cos(\theta)v - \sin(\theta)u\right]\right)\tag{2}\\ &= \frac{\partial}{\partial \theta}\left(\frac{1}{r}\left[\sin(\theta)v + \cos(\theta)u\right] + i \frac{1}{r}\left[\cos(\theta)v - \sin(\theta)u\right]\right)\tag{3} \end{align} Note: These last equations can be obtained from the polar C-R equations as in this answer.
After this, I tried combining equations $(1), (2)$ and $(3)$ to get a system of differential equations from where I could solve for $u$ and $v$ explicitly, and therefore solving for $f$ since $f = u + iv$.
The problem I encountered was that I couldn't solve for the values of the constants of integration. As much as I tried I just kept getting expressions that I couldn't simplify.
Is my approach correct? Or alternatively, does anyone know a simpler method I could use to prove this? Thank you!