Complex convergence of sequence Proof check and help Prove that, for $z \in C$ , the sequence $(z^{n})$
 converges if and only if $|z| < 1$ or $z=1$
Proof. Say $z^n$ converges to some $a$ then there exists $n_o$ such that for all $n\geq n_o$
$|z^n-a|\leq ε$ for every $ε\geq 0$ now $$|z^n|-|a| \leq |z^n-a| \leq ε$$ but if $|z| > 1$ then $|z^n|-|a|$ does not converge since  $|z^n|-|a| \leq ε$
$$|z^n|- \leq ε+|a|$$ => $|z|^n \leq ε+|a|$ => $nlog{|z|} \leq log(ε+|a|)$ 
Because $log|z| > 0$  i get  $$n \leq \frac{log(ε+|a|)}{log|z|} $$ for every $n\geq n_0$ which is not true for every $ε$ because $n \geq integerpart (\frac{log(ε+|a|)}{log|z|})+1$ .
So $|z| \leq 1$ .
Now for the second part. Suppose $|z|\leq1$ this means $$|z|^n \leq z^{n-1} \leq |z| \leq1$$ Have to prove that $$|z^n-a| \leq ε$$ for every $ε \geq 0$ for every $n \geq n_o$  I tried to write z in its polar form  $$z^n=|z^n|e^{inΘ}$$ split it imaginary part and real take the modulus $$|z^n-a| \leq ε=>||z^n|e^{iθn}-a| \leqε $$ try and solve for $n$ but didnt got me to anywhere.
 A: If $\lvert z \rvert < 1$, then $\lvert z^n \rvert = \lvert z \rvert^n \to 0$, which is equivalent to $z^n \to 0$. On the other hand, if $\vert z \rvert > 1$, then $\lvert z^n \rvert = \lvert z \rvert^n \to \infty$, so the sequence $(z^n)$ cannot converge.
Now for the behavior on the unit circle. Convergence at $z = 1$ is clear. Suppose that $\lvert z \rvert = 1$ and $z \neq 1$. If the argument of $z$ is a rational multiple of $\pi$, then $(z^n)$ traces a periodic orbit on the circle, hence it cannot converge. If on the other hand the argument of $z$ is an irrational multiple of $\pi$, then the image of $(z^n)$ is dense in the unit circle (see e.g. this question: Dense set in the unit circle- reference needed). If the sequence were to converge to some number $z^\infty$, then the set $\{z^n\} \cup \{z^\infty\}$ would be closed in the circle $S^1$. But then, by density, $$\{z^n\} \cup \{z^\infty\} = \overline{\{z^n\} \cup \{z^\infty\}} = S^1,$$
where the bar denotes topological closure. This is a contradiction as the left-hand side is countable and the right-hand side isn't.
