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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.

Jam
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    The result you're trying to prove isn't true. In fact, $(z^n)$ converges iff $|z|<1$ or $z=1$. – Eric Wofsey Jul 23 '16 at 15:58
  • @EricWofsey thats what im trying to prove. it says iff $ |z| \leq 1$ at the start of the problem i wrote it at the start of the answer. – Jam Jul 23 '16 at 16:01
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    $|z|<1$ or $z=1$ is NOT the same as $|z|\leq 1$. – Lukas Betz Jul 23 '16 at 16:04
  • @EricWofsey ohhh haha my eyes did not see there was no modulus .you are right – Jam Jul 23 '16 at 16:10
  • The cases $|z|\lt 1$ and $|z|\gt 1$ can be handled by norm calculations. A different argument will be needed to show that if $|z|=1$ but $z\ne 1$, then $z^n$ does not converge. – André Nicolas Jul 23 '16 at 16:52

1 Answers1

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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.

Community
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Alex Provost
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  • Damn my whole thinking was so bad. One question though i get that z^n set is dense. and thus the closure of it will have to be the same as the unit circle . But dont i have to get the contradiction by saying that the sequence converges.I mean even if i didnt say that the sequence converges i would still get a contradtiction because the $z^n$ is countable dense and its closure must be equal to the Unit circle. hence contradiction without adding the extra convergance point. – Jam Jul 23 '16 at 19:12
  • " If the argument of z is rational" That should be "rational multiple of $\pi$" – zhw. Jul 23 '16 at 19:14
  • @zhw. Yes of course, thank you. The identification $S^1 \cong \mathbb{R}/\mathbb{Z}$ is too ingrained in my head! – Alex Provost Jul 23 '16 at 19:18
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    @ManolisLyviakis Well you want to relate the image of the sequence to its closure in some way to apply this argument. If the sequence doesn't converge then it's not guaranteed to contain all its adherent points. You would be able to say that ${ z^n} \subseteq \overline{{z^n}} = S^1$, but that's not a contradiction. – Alex Provost Jul 23 '16 at 19:26
  • If i take rational multiples of pi why i know it wont converge? – Jam Jul 23 '16 at 19:29
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    @ManolisLyviakis If the argument of $z$ has the form $\pi p/q$, then the argument of the sequence $(z^n)$ cyclically takes on the values $\pi p/q, 2\pi p/q, \ldots, 2q\pi p/q = 2\pi p = 0, \pi p/q, \ldots$. In other words, the image of the sequence forever and cyclically visits a finite set of points. – Alex Provost Jul 23 '16 at 19:37
  • Cant find a definition of density inside a set only on a topological space.Is $S_1$ a topological space? – Jam Jul 25 '16 at 12:42
  • @ManolisLyviakis Yes, the circle is a topological space with the subspace topology induced from the complex plane. One even needs a topology to speak of convergent sequences. Everything you do in analysis depends crucially on the underlying topologies. – Alex Provost Jul 25 '16 at 15:06