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Let $f : \mathbb{C} \to \mathbb{C}, f(z) = z^3$.

We say that $f$ is expanding on the unit sphere $S^1$ if there is $c > 1$ and $\varepsilon > 0$ such that for every $z, w \in S^1$, if $d(z, w) < \varepsilon$, then $d(f(z), f(w)) > c d(z, w)$.

I'm trying to show that $f$ is expanding on $S^1$ but I can't. Can someone give me a hint, please?

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  • $\begingroup$ What distance function are you using? Also, it suffices to prove this for $z=1$ as any sensible distance is invariant under rotation/translation (choose one depending on the way you choose to view $S^1$.) $\endgroup$ – Servaes Jan 13 at 13:07
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$f'(1) = 3$, therefore there is an $\epsilon > 0$ such that $$ 0 < |\zeta-1| < \epsilon \implies \left | \frac{\zeta^3 - 1}{\zeta-1} \right| > 2 \, . $$

Now assume that $z, w \in S^1$ with $0 <|z-w| < \epsilon$. Without loss of generality, $w \ne 0$. Applying the above estimate to $\zeta = z/w$ gives the desired conclusion.

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  • $\begingroup$ Can you explain me why $$ 0 < |\zeta-1| < \epsilon \implies \left | \frac{\zeta^3 - 1}{\zeta-1} \right| > 2 \, , $$ please? $\endgroup$ – g.pomegranate Jan 14 at 11:03
  • $\begingroup$ @g.pomegranate: $3 = | f'(1) | = \lim_{\zeta \to 1} \left| \frac{\zeta^3 - 1}{\zeta-1} \right |$, therefore the fraction on the right-hand side is close to $3$ if $\zeta$ is sufficiently close to $1$. $\endgroup$ – Martin R Jan 14 at 12:02
  • $\begingroup$ Thank you very much! $\endgroup$ – g.pomegranate Jan 14 at 13:04

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