# Prove that if the $n$-th derivative of analytic function is $0$, then it's a polynomial. [duplicate]

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!

• This is a very complicated way of proceeding. If you know the result is true for some $n$, then $f^{(n+1)}=0$ implies $f'(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}$. Therefore $f(z)=h(z)+a_0z+a_1z^2/2+\cdots+a_{n-1}z^n/n$ where $h'(z)=0$. So all you need is that functions with zero derivative are constant. Commented Aug 16, 2020 at 19:08
• @AnginaSeng, my problem is that I don't know how to justify going from $f'(z)$ to $f(z)$ in complex functions. Every problem I've seen up to now when dealing with derivatives comes down to simplifying it into a problem of derivatives of real functions by using Cauchy-Riemann. I haven't seen the concept of a complex antiderivative, so this is why I was taking this convoluted approach. Commented Aug 16, 2020 at 19:13
• All you need is that (i) the derivative of $z^n$ is $nz^{n-1}$ and (ii) a function on a domain with zero derivative is constant. Commented Aug 16, 2020 at 19:14
• Ohhhh, I see. That is a way simpler approach. I think I can easily prove those 2 conditions. Thank you! Commented Aug 16, 2020 at 19:17