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Exactly what it says on the tin.

Find for what values of $z\in \mathbb{C}$ does this limit exist: $$\lim_{n\to\infty} z^n - z^{2n}$$

The cases when $|z|\not = 1$ are fairly simple, but I am struggling with $|z|=1$.

My intuition and wolfram alpha tell me that it diverges if $z\not = 1$, but I cannot formalize the argument.

Jsevillamol
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  • Related: http://math.stackexchange.com/questions/238997/prove-the-divergence-of-the-sequence-left-sinn-right-n-1-infty – Spenser Oct 07 '16 at 12:27

2 Answers2

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Hint: Well, $z^n - z^{2n}$ has a limit as $n$ goes to $+ \infty$ if and only if (why?) $\mathfrak{Re}(z^n - z^{2n})$ and $\mathfrak{Im}(z^n - z^{2n})$ have both a limit as $n$ goes to $+ \infty$. If $|z| = 1$, then $z = e^{i\theta}$. Therefore, study the limit of $x_n =\cos(\theta n)- \cos(2\theta n)$ and $y_n = \sin(\theta n)- \sin(2\theta n)$ as $n$ goes to $+ \infty$ given the value of $\theta$

Hermès
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Ok, nevermind.

If we suppose that $\lim_{n\to\infty} z^n - z^{2n}=\lambda$ then it should happen that $\lim_{n\to\infty} (z^n - z^{2n}) - (z^{n+1} - z^{2n+2}) = 0$.

But $$ (z^n - z^{2n}) - (z^{n+1} - z^{2n+2}) = z^n (1-z) (1-z^{2n}(1+z) $$ taking the module, $|z^n (1-z) (1-z^{2n}(1+z)|> |1-z|(|2|+|z|) > 0$ (since $z\not=1$), so contradiction and GG.

Jsevillamol
  • 4,668