Show that $\sum_{k=1}^{n}a_ke^{2 \pi ikx}$ has a root in $\left[ 0,1 \right]$

Question

Let $a_1, \dots ,a_n$ be arbitrary complex numbers. Define: $f \left( x \right)=\sum_{k=1}^{n}a_ke^{2 \pi ikx}$. I wish to show that there exists an $x\in\left[ 0,1 \right]$ s.t. $f \left( x \right)=0$.

What I tried:

It's pretty straightforward to show that $\int_0^1e^{2 \pi ikx}\mathrm{d}x=0 $ ($k\geq1$) and thus $\int_0^1f \left( x \right)\mathrm{d}x=\int_0^1\sum_{k=1}^{n}a_ke^{2 \pi ikx}\mathrm{d}x=0$. Now if we had the equivalent of the mean value theorem in complex analysis we'd be done. I tried to think of some way to apply the mean value property to yield similar results but I can't see how it might help.

Answer 1

Take $a_1 = 1$, $a_i = 0 \ \ \forall i\geq 2$ then $$f(x) = e^{2 \pi i x}$$ that is never zero.

Answer 2

If $w = e^{2 \pi i x}$, your $f(x) = \sum_{k=1}^n a_k w^k = w g(w)$ where $g(w) = \sum_{k=1}^n a_k w^{k-1}$ is an arbitrary polynomial of degree $ \le n-1$. Unless all $a_k = 0$, this has $n-1$ roots (counted by multiplicity) in the complex plane. Of course roots $w=0$ don't translate to solutions of $f(x)=0$.

The only cases where the only roots of $g(w)$ are at $w=0$ are where $g(w) = c w^{j}$ for some constant $c$, i.e. $f(x)$ has only one nonzero term. Otherwise, $f(x) = 0$ will always have some solution in the complex numbers. However, there is no reason to expect a real solution, unless you have more information.