Prove for $\epsilon>0$ there are infinite $k$ so $\sqrt[k]{|z_1^k+z_2^k+\cdots+z_n^k|}>\max(|z_1|,|z_2|,...,|z_n|)-\epsilon$

Question

Let $z_1, z_2,...,z_n$ be arbitrary complex numbers. prove that for any $\epsilon > 0$ there are infinitely many numbers $k$ such that \begin{equation} \sqrt[k]{|z_1^k + z_2^k + \cdots + z_n^k|} > \max(|z_1|, |z_2|,...,|z_n|) - \epsilon \end{equation}

One idea i had to prove this was to do a proof by contradiction. That is, suppose $\exists \epsilon, \exists N$ such that $k > N \implies \sqrt[k]{|z_1^k+z_2^k+\cdots+z_n^k|}<\max(|z_1|,|z_2|,...,|z_n|)-\epsilon$. We can suppose $|z_1| = \max(|z_1|,|z_2|,...,|z_n|)$ so that dividing both sides by $|z_1|^k$ we get \begin{equation} |1+S_k| < (1 - \epsilon)^k \end{equation}

where $S_k = \frac{z_2^k + \cdots+ z_n^k}{z_1^k}$. I was thinking that letting $k \to \infty$ should give a contradiction, but I'm not sure how. Perhaps a helpful hint is that I'm supposed to use formal power series to solve this.

Answer 1

Consider the function $$f(x)=\sum_{i=1}^n \frac1{1-z_ix}=\sum_{m=0}^\infty x^m(z_1^m+\cdots+z_n^m).$$ This function has a pole at $1/z_i$ for every $i$. However, if $$\limsup_{m\to\infty}\left|z_1^m+\cdots+z_n^m\right|^{1/m}=s,$$ then the second series definition for $f(x)$ converges for all $x$ with $|x|<1/s$. If this is strictly less than $\max(|z_1|,\dots,|z_n|)$, then this contradicts the existence of a pole at $1/z_i$ for whichever $i$ maximizes (not necessarily uniquely) $|z_i|$, which is inside the radius of convergence.

Answer 2

As the simple visualization in $\mathbb R$ shows this fails because the $max$ function touches the k-root of the sum of n arbitrary z complex numbers at certain directions and in this, both are usually equal or identic. In the case of the square root and the max function, these are the axes.

I show these facts only for positive reals:

This is the case for $k=0.2$

This is the case for $k=5$

Mind that the number of samples $z_{i}$ is free for the proof.

All values zero is the trivial case. The functions are steady on $\mathbb R$ The monotony argument holds.

Have a look at the difference function in the last example:

This is strictly positive for positive reals. So Your left-hand site is always bigger than the other one with $\epsilon==0$. $\epsilon$ is simply not needed for inequality to hold and can not be other than trivial on the coordinate system axes. The no inequality holds if $\epsilon==0$.

$\epsilon<0$ constant implies that the angle is reduced and the linear boundary gets curved. The inequalities still holds. Since both compared functions are homogenous of the linear order. Multiple the same $z_{i}$ do not change anything. The right-hand site is not additive as the left-hand site is somehow if more numbers are added. This increase the inequality that holds without subtracting something from the $max$ - function.