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 ?

  • $\begingroup$ $e^{log(z)/2} = e^{log(z^{1/2})} = z^{1/2} $? $\endgroup$ – Displayname Feb 10 at 15:15
  • $\begingroup$ @ 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 $\endgroup$ – 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 $\left(e^{\frac12\log z}\right)^2=e^{\log z}=z$.

  • $\begingroup$ 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,$$(\forall z\in U):e^{\log z}=z.$$ $\endgroup$ – José Carlos Santos Feb 10 at 16:09

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