Let $\Omega\subseteq\mathbb{C}$ a region $\mathcal{L}:\Omega\to\mathbb{C}$ a logarithm function and $f:\Omega\to\mathbb{C}$ a continuous function such that $(f(z))^n = z$ for all $z\in\Omega$. Prove that there is $0\leq k\leq n-1$ such that $f(z)=e^{\frac{2\pi ik}{n}}e^{\frac{1}{n}\mathcal{L}(z)}$ for all $z\in\Omega$
Idea: if $\mathcal{L}:\Omega\to\mathbb{C}$ is a logarithm function then $e^{\mathcal{L}(z)}=z$ for all $z\in\Omega$. Note that: $(f(z))^n=z=(e^{{2\pi i}}e^{\frac{1}{n}\mathcal{L}(z)})^n=e^{{2\pi in}}e^{\mathcal{L}(z)}$ But I don't know how to continue, can anyone help me? Please
Thanks
Define $g(z)=f(z)\cdot e^{-\frac{1}{n}\mathcal{L}(z)}$, $(g(z))^n=1$ for all $z\in\Omega$.
Then $g(z)$ it is a root of unity, therefore it exists $k\in{\{0,1,2,...,n-1}\}$ such that $g(z)=e^{\frac{2\pi ik}{n}}$ for all $z\in\Omega$.
Then $g(z)=f(z)\cdot e^{-\frac{1}{n}\mathcal{L}(z)}=e^{\frac{2\pi ik}{n}}$. Hence $f(z)= e^{\frac{2\pi ik}{n}}e^{\frac{1}{n}\mathcal{L}(z)}$ for all $z\in\Omega$.