# Showing that $z \to z^3$ is expanding on the unit sphere

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?

• 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 $S^1$.) – Servaes Jan 13 at 13:07

## 1 Answer

$$f'(1) = 3$$, 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.

• Can you explain me why $$0 < |\zeta-1| < \epsilon \implies \left | \frac{\zeta^3 - 1}{\zeta-1} \right| > 2 \, ,$$ please? – g.pomegranate Jan 14 at 11:03
• @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 $1$. – Martin R Jan 14 at 12:02
• Thank you very much! – g.pomegranate Jan 14 at 13:04