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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!

Robert Lee
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  • 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. – Angina Seng Aug 16 '20 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. – Robert Lee Aug 16 '20 at 19:13
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    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. – Angina Seng Aug 16 '20 at 19:14
  • Ohhhh, I see. That is a way simpler approach. I think I can easily prove those 2 conditions. Thank you! – Robert Lee Aug 16 '20 at 19:17

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