Proving $\rvert 1+\cos(z)+\cos(2z)+ \dotsb +\cos(nz)\lvert>n+1$

Question

Let $n\in \Bbb N$. Show that for each $r>0$ there exists $z\in D(0,r)$ such that $\rvert 1+\cos(z)+\cos(2z)+ \dotsb +\cos(nz)\lvert>n+1$.

I think it could work with the Maximum Principle theorem, but I'm not sure about how is that function bounded on $D(0,r)$.

Answer 1

Your function $f_n(z)=1+\cos z+\cos 2z+\dots + \cos nz$ is entire. It is bounded on any $D(0,r)$ as any continuous function on bounded domains is bounded.

$f_n$ furthermore satisfies $f_n(0)=n+1$. If $|f_n(z)|\le n+1$ everywhere in $D(0,r)$, that means $0$ is an interior maximum, so $f_n$ must be constant by the maximum principle. As $f_n$ is not constant, the bound must be wrong.

For a proof using significantly less technology, you might want to consider $f_n(ir/2)$ or something similar, and you should be able to show directly that $|f_n(ir/2)|>n+1$.

Answer 2

The function $f(z):=\sum_{k=0}^n \cos(k\,z)$ is noncnstant. It therefore maps any neighborhood $D(0,r)$ of $0$ onto a neighborhood $V$ of $f(0)=n+1$. Such a $V$ contains necessarily points of absolute value $>n+1$.