Show that the angle of rotation at $z_0 = r_0 e^{i\theta _0 }$ under $w = z^n$ is $(n-1)\theta _0$.

Question

Show that the angle of rotation at a nonzero point $z_0 = r_0 e^{i\theta _0 } \in \mathbb{C}$ under the transformation $w = z^n (n = 1, 2, \cdots )$ is $(n-1)\theta _0$ . Show that the scale factor of the transformation at that point is $nr_0 ^{n-1}.$

I know that multiplying complex numbers causes rotation, and that we can calculate the new coordinates on the circle of radius $r_0$, where the points are separated by $\theta _0$ (change the exponential into $\cos \theta + i \sin \theta$ form so that the arguments add as usual since we are multiplying), but I am not sure how to do this with $z^n$ to get the result we want.

Answer 1

Note that $f$ is analytic at $z_0$ and $f’(z_0) \neq 0 $, so $f(z)$ is conformal at $z_0$. The angle of rotation $\psi$ is given by $\psi = \text{arg } f’(z_0)$ and the scale factor $\lambda$ is given by $\lambda = |f’(z_0)|$. So, $$\psi = \text{arg } nr_0 ^{n-1} \exp[i\theta_0(n-1)] = \theta _0 (n-1)$$ and $$ \lambda = \left| nr_0 ^{n-1} \exp[i\theta_0(n-1)] \right| = nr_0 ^{n-1}.$$