Does $\lim_{n\to+\infty}\sqrt[n]{z_1^{n}+z_2^{n}+...+z_k^{n}}=0\Rightarrow z_1=z_2=...=z_k=0$ hold for complex numbers? $$\lim_{n\to+\infty}\sqrt[n]{z_1^{n}+z_2^{n}+...+z_k^{n}}=0\Rightarrow z_1=z_2=...=z_k=0$$
This is right for real numbers because if we let $z_i=\max\{z_j\}$,
$$0=\lim_{n\to+\infty}\sqrt[n]{z_1^{n}+z_2^{n}+...+z_k^{n}}=\lim_{n\to+\infty}z_i\sqrt[n]{(\frac{z_1}{z_i})^{n}+(\frac{z_2}{z_i})^{n}+...+(\frac{z_i}{z_i})^{n}+...+(\frac{z_k}{z_i})^{n}}=z_i$$
$$\Rightarrow \max\{z_j\}=0\Rightarrow z_j=0,\ \ \forall j\in\{1,2,3,...k\}$$
My question is, does this hold for all complex numbers? Since in this case $\sqrt[n]{x}$ may have multiple values, we will only use one of them, which makes the thing more complicate.

Background: Let M be a $k\times k$ matrix, if $trace(M^n)=0,\ \ \forall n\in \mathbb{N}$, then M is nilpotent.
We can proof this with Newton's identity or Vandermonde determinant. But they are both algebraic methods. If the proposition above is true, we can find an analytical way.
 A: The result is true and I will sketch the proof - the only extra ingredient is the fact that if $|w_r|=1, r=1,..m$ then $w_1^n+...w_m^n$ cannot converge to zero as $n \to \infty$ which follows easily from say newton formulas since $a_1...a_m$ can be expressed polynomially in terms of the sums $a_1^r+...+a_m^r, r=1,.m$ with coefficients independent of the actual values of $a_r$ so assuming all those sums are at most $\epsilon >0$ in absolute values, we can get $a_1...a_m$ at most $1/2$ say in absolute value for $\epsilon$ small enough depending only on $m$. But now if by contradiction $w_1^n+...w_m^n$ converges to zero, we apply this to $a_r=w_r^n$ for large enough $n$ so all the sums above are at most $\epsilon$ so $1=|w_1^n...w_m^n|<1/2$ contradiction
Now we repeat the argument from the real case and if at least one $z_k \ne 0$ we divide by the maximum of $ |z_r|$ and get $|1^n+w_1^n+...w_m^n| \to 0$ with $|w_k| \le 1$; we can clearly ignore those with modulus strictly less than $1$ as then $w_r^n \to 0$ so we can assume $|w_r|=1$ and the above proof gives the required contradiction
A: We will use proof by contradiction.
Without loss of generality, we may assume that each $|z_{j}| \leq 1$ and $\max_{1 \leq j \leq k} |z_{j}| = 1$ (i.e by pulling out $\max_{1 \leq j \leq k}|z_{j}|$ from your above equation and considering the new limit.). We can also without loss of generality arrange $0<|z_{1}| \leq |z_{2}| \leq ... \leq |z_{k}| = 1$.
By Simultaneous Dirichlet's approximation theorem (https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem), for each $N \geq 1$ we can choose $1 \leq N_{q} \leq N$ and integers $a_{q,1},...,a_{q,k}$ so that $$|\frac{\arg(z_{j})}{2\pi} - \frac{a_{q,j}}{N_{q}}| \leq \min \{\frac{1}{N_{q}N^{\frac{1}{k}}}, \frac{1}{2N_{q}}\}$$
for each $1 \leq j \leq k$. Note that for each $c \in \mathbb{N}$ we have;
$$|z_{j}^{c N_{q}} - |z_{j}|^{c N_{q}}|$$
$$ = |z_{j}|^{cN_{q}}|e^{icq\arg(z_{j})} - 1|$$
$$= |z_{j}|^{cN_{q}}|e^{ic\theta_{j}} - 1|$$
where $|\theta_{j}| \leq \frac{2\pi}{N^{\frac{1}{k}}},$ continuing the above chain, the above quantity is;
$$= |z_{j}^{N_{q}}||2\sin(\frac{c\theta_{j}}{2})|$$
$$\leq |z_{j}^{N_{q}}|(\frac{2\pi c}{N^{\frac{1}{k}}})$$
$$\leq (\frac{2\pi c}{N^{\frac{1}{k}}}). $$
Thus $|\sum_{j=1}^{k}z_{j}^{cN_{q}} -|z_{k}|^{cN_{q}}| \leq \frac{2\pi k c}{N^{\frac{1}{k}}}$
Thus
$$\sqrt[cN_{q}]{\sum_{j=1}^{k} z_{j}^{cN_{q}}} = \sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{cN_{q}} + V(N,k,c)}$$
Where $1 \leq \sum_{j=1}^{k} |z_{j}|^{cN_{q}} \leq k$ and $|V(N,k,c)| \leq \frac{2\pi k c}{N^{\frac{1}{k}}}$
Now observe that
$$V(N,k, c) = \left(\sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{c N_{q}} + V(N,k,c)}\right)^{c N_{q}} - \left(\sqrt[c N_{q}]{\sum_{j=1}^{k} |z_{j}|^{c N_{q}}}\right)^{c N_{q}} $$
$$= \left(\sqrt[c N_{q}]{\sum_{j=1}^{k} |z_{j}|^{c N_{q}} + V(N,k,c)} - \sqrt[c N_{q}]{\sum_{j=1}^{k} |z_{j}|^{c N_{q}}}\right)\left( \sum_{h = 0}^{cN_{q}-1}\left(\sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{cN_{q}} + V(N,k,c)}\right)^{h}\left(\sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{cN_{q}}} \right)^{cN_{q}-1-h}\right)$$
Thus $$|V(N,k,c)| \geq \left|\left(\sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{cN_{q}} + V(N,k,c)} - \sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{cN_{q}}}\right)\right|\left|\left(\sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{cN_{q}}} \right)^{cN_{q}-1} \right|$$
$$\Rightarrow |V(N,k,c)| \geq \left|\left(\sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{cN_{q}} + V(N,k,c)} - \sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{cN_{q}}}\right)\right|$$
$$\Rightarrow \left|\left(\sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{cN_{q}} + V(N,k,c)} - \sqrt[cN_{q}]{\sum_{j=1}^{k} |z_{j}|^{cN_{q}}}\right)\right| \leq \frac{2\pi kc}{N^{\frac{1}{k}}} $$
To finish the proof there are two cases;
$\textbf{Case 1)}$ The Sequence $N_{1}, N_{2},...$ contains infinitely many distinct  positive integers.
In this case, suppose the subsequence of distinct positive integers is $n_{1}<n_{2}<...$, from the above inequality we have
$$\left|\sqrt[n_{j}]{\sum_{d=1}^{k} |z_{d}|^{n_{j}}}\right| \leq \left|\sqrt[n_{j}]{\sum_{d=1}^{k} z_{d}^{n_{j}}}\right| + \frac{2\pi k}{n_{j}^{\frac{1}{k}}}$$
Thus
$$\lim_{j \rightarrow \infty} \sqrt[n_{j}]{\sum_{d=1}^{k} |z_{d}|^{n_{j}}} = 0$$
This implies that $|z_{1}| = |z_{2}| = ... |z_{k}| = 0$, this is impossible.
$\textbf{Case 2)}$ There exists $t \in \mathbb{N}$ so that $N_{t} = N_{t+1} = ...$. In this case it is obvious that $$\arg(z_{j}) = \frac{2 \pi a_{t,j}}{N_{t}}$$
Hence for each $c \in \mathbb{N}$ one has that
$$\sqrt[cN_{t}]{\sum_{j=1}^{k}z_{j}^{cN_{t}}} = \sqrt[ct]{\sum_{j=1}^{k}
|z_{j}|^{cN_{t}}}$$
Thus $\lim_{c \rightarrow \infty} \sqrt[ct]{\sum_{j=1}^{k}
|z_{j}|^{cN_{t}}} = 0$
This would imply $|z_{1}| = ... = |z_{k}| = 0$, which is impossible.
Since the above two cases are impossible we have a contradiction.
