Let $z_n$ be a complex sequence. The limit $\lim_{n\rightarrow\infty}z^3_n$ exists if and only if the limit $\lim_{n\rightarrow\infty}z_n$ exists

I think the statement is true. From the definition, an infinite sequence of complex numbers has a limit $z$ if, for each positive number $ε$, there exists a positive integer $n_0$ such that $|z_n − z|<ε$ whenever $n>n_0$.

How can I use the definition to prove the statement above? Is it true in the first place?

  • 1
    $\begingroup$ If $\zeta$ is a primitive third root of $1$ and $z_n= \zeta^{n \mod 3}$, then $z_n$ doesn't converge, but $z_n^3$ is constant. $\endgroup$ Apr 15, 2021 at 22:27

1 Answer 1


The statement is false. Let $\omega=-\frac12+\frac{\sqrt3}2i=e^{2\pi i/3}$, and let $z_n=\omega^n$. Then $(\forall n\in\Bbb N):z_n^{\,3}=1$, but the sequence $(z_n)_{n\in\Bbb N}$ diverges.

  • $\begingroup$ Can this counter example be used to disprove the opposite statement as well? ($\lim_{n\rightarrow\infty}z^3_n = inf$ if and only if $\lim_{n\rightarrow\infty}z_n = inf$ ) $\endgroup$
    – Y.I.
    Apr 15, 2021 at 22:38
  • $\begingroup$ What is $inf$? Do you mean $\infty$? $\endgroup$ Apr 15, 2021 at 22:38
  • $\begingroup$ Yes I mean $\infty$ $\endgroup$
    – Y.I.
    Apr 15, 2021 at 22:39
  • $\begingroup$ No. It is actually true that $\lim_{n\to\infty}z_n^{\,3}=\infty\iff\lim_{n\to\infty}z_n=\infty$. $\endgroup$ Apr 15, 2021 at 22:41
  • $\begingroup$ Thank you, I will try to think of the proof for this. $\endgroup$
    – Y.I.
    Apr 15, 2021 at 22:42

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