Be $\rho > 0, \alpha < \beta$, with $\rho\alpha,\rho\beta,\alpha,\beta \in [-\pi,\pi]$. Prove that $z \rightarrow z^{\rho}$ is a bijection of $\Omega_1 = \{z \in \mathbb{C^*}: \alpha < arg \ z < \beta \}$ in $\Omega_2 = \{z \in \mathbb{C^*}: \rho\alpha < arg \ z < \rho\beta \}$
What I've thought is to work with the exponential $e^{\rho \ log \ z}$ instead of $z^{\rho}$ but I don't find the manner of proving there is a bijection. Cound you take me a hand?
PD: Could you help too with this problem? Mapping exponential function to complex plane
It's just the map $f(re^{i\theta}) = r^{\rho}e^{i\rho\theta}$, with inverse $g(re^{i\theta}) = r^{1 \over \rho}e^{i{\theta \over \rho}}$.