I'm looking at a question that asks me to show that:
If a function $f$ is known to be $n$-times differentiable in a domain $D$ and if $\forall{z\in{D}}\ \ f^{(n)}(z)=0$, then $f$ is a polynomial function of $z$ whose degree is at most $n-1$.
There is a hint to use induction.
My concern is that, in order to do this, I believe I must first show that $f$ is a polynomial and no other function, otherwise I won't be able to use the inductive step that I believe I should take.
Proceeding as such, I can write $f(z)=u(z)+iv(z)$, so $f^{(n)}(z)=u_{x^{n}}(x,y)+iv_{x^n}(x,y)=0$ which means that $u_{x^n}(x,y)=u_{y^n}(x,y)=0$ and $v_{x^n}(x,y)=v_{y^n}(x,y)=0$ and the equality of the partials given by the Cauchy-Riemann Equations follow because of the differentiability of $0$. If this were just calculus and this function were restricted to $\mathbb{R^2}$, then I would be able to integrate the $u$ and $v$ here to demonstrate that it is a polynomial, but since this is restricted to a domain $D$ and I'm dealing a complex function, I'm not sure if I can still do that.
If I consider the Cauchy-Riemann Equations, then I don't know if I can integrate the real valued functions $u$ and $v$ given by $f(x,y)=u(x,y)+iv(x,y)$
Is there something I'm missing here?
The hard part is the base case, $n=1$:
Once you have this, the rest is an easy induction.
Take $g=f'$. Then $g^{(n-1)}=0$ in $D$. By induction, $g$ is a polynomial of degree at most $n-2$. Integrate $g$ formally and get a polynomial $G$ of degree at most $n-1$ with $G'=g$. Let $h=G-f$. Then $h'=g-f'=0$ and so $h$ is a constant, by the base case. Therefore, $f=G+h$ is a polynomial of degree at most $n-1$.
The base case can be proved using the Cauchy–Riemann equations and the fact that $D$ is connected by paths, even staircase paths. See this proof for instance.