# Some confusion about complex logarithm on simple connected set (In the proof of Riemann mapping theorem) .

Let $$D$$ denote the unit disc , and $$U$$ is an open simple connected subset of $$D-\{0\}$$ , then we can define a square root function on $$U$$ by $$g(z)=e^{\frac12 \log z}$$ such that $$g$$ is an injective holomorphic mapping from $$U \to D$$

The statement above was in Stein's complex analysis Page $$_{230}$$ and I'm quite confused about the function $$\log z$$ defined here . Indeed , what we need here is $$z^{\frac12}$$ and we expect that whenever $$z=re^{i \theta}$$ we can have $$|z^{\frac12}|=r^{\frac12}$$ here . However , since I only know $$U$$ is simple connected , how could I define $$g$$ here ?

• $e^{log(z)/2} = e^{log(z^{1/2})} = z^{1/2}$? – Displayname Feb 10 at 15:15
• @ Displayname Yes , that was the definition of $z^{\frac12}$ here . Since $z$ is a complex variable , we can not define it as the real variable function – J.Guo Feb 10 at 15:20

I am assuming that $$0\notin U$$. In that case, since $$U$$ is simply connected, $$\frac1z$$ has a primitive $$\psi$$ in $$U$$. Take $$z_0\in U$$ and choose $$w_0\in\mathbb C$$ such that $$\psi(z_0)+w_0$$ is a logarithm of $$z_0$$. Now, let $$\log z$$ be $$\psi(z)+w_0$$. Then, for each $$z\in U$$, $$\log z$$ is a logarithm of $$z$$. So $$\left(e^{\frac12\log z}\right)^2=e^{\log z}=z$$.
• No, we don't need to have $1\in U$. Consider the map $\log$ that I defined in my answer. Let $h(z)=\frac{e^{\log z}}z$. Then$$h'(z)=\frac{z\log'(z)e^{\log z}-e^{\log z}}{z^2}=\frac{e^{\log z}-e^{\log z}}{z^2}=0.$$So, since $U$ is connected, $h$ is constant. But $h(z_0)=\frac{z_0}{z_0}=1$, and therefore $h(z)$ is always $1$. In other words,$$(\forall z\in U):e^{\log z}=z.$$ – José Carlos Santos Feb 10 at 16:09