Conformality of $z^n$

Question

For $n\gt 1$, the function $z^n$ multiplies angles at the origin by $n$, so it is not conformal at $z=0$. The following theorem shows that $z^n$ is conformal at any other point $z$ other than $0$.

Theorem. If $f(z)$ is analytic at $z_0$ and $f'(z_0)\neq0$, then $f(z)$ is conformal at $z_0$.

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Why $z^n$ is not conformal at the origin? I agree with the theorem that $f'(z_0)\neq0$, but I would like to see a "geometric" intuition why this is not true?

Answer 1

Yes, it is. "Conformal" means preserving the angles. The mapping $z \mapsto z^{n}$ changes the angles at the origin.

Answer 2

Note that the ray $\gamma_t = \{ re^{it} \mid r \ge 0 \}$ is mapped to $\gamma_{nt} = \{ re^{nit} \mid r \ge 0 \}$, by $f(z) = z^n$.

In particular, the angle between $\gamma_0$ and $\gamma_t$ is multiplied by $n$ under the mapping $f(z) = z^n$, so it is not conformal at $z=0$.