- Let f be a one-to-one analytic function defined on the unit disk D centered at the origin and \(f ( 0 ) = 0\) . Show that the function \(g ( z ) = { \sqrt { f ( z ^ { 2 } ) } }\) has a single-valued branch and is also one-to-one.
Solution
1. There is a holomorphic nonvanishing function \(h\) on \(D\) with \(f(w)=wh(w)\).
The Taylor series of \(f\) at zero has zero constant term, so \(f(w)/w\) extends holomorphically across zero with value \(h(0)=f'(0)\). An injective holomorphic function has nonzero derivative [@SS03], hence \(h(0)\ne0\). For \(w\ne0\), injectivity and \(f(0)=0\) give \(f(w)\ne0\), so \(h(w)\ne0\). Thus \(h\) has no zero anywhere in \(D\).
2. An explicit odd holomorphic square root exists on the whole disk.
Since \(h'/h\) is holomorphic on the simply connected disk, it has a holomorphic primitive \(L_0\) with \(L_0(0)=0\) [@SS03]. Choose \(c\in\mathbb C\) with \(e^c=h(0)\) and set \(L=L_0+c\). Differentiation gives \[ (he^{-L})'=e^{-L}(h'-hL')=0. \] Its value at zero is one, so connectedness implies \(h=e^L\) on \(D\). Define \[ g(z)=z\exp\left(\frac12L(z^2)\right),\qquad z\in D. \] This is a single-valued holomorphic function, and \[ g(z)^2=z^2h(z^2)=f(z^2). \] It also satisfies \(g(-z)=-g(z)\) and has no zero except at \(z=0\), because the exponential never vanishes.
3. The branch \(g\) is injective.
Suppose \(g(z_1)=g(z_2)\). Squaring and using step 2 give \(f(z_1^2)=f(z_2^2)\). Injectivity of \(f\) yields \(z_1^2=z_2^2\), so \(z_2=z_1\) or \(z_2=-z_1\). In the second case, oddness gives \(g(z_1)=g(-z_1)=-g(z_1)\), whence \(g(z_1)=0\). Step 2 then forces \(z_1=0=z_2\). Thus in every case \(z_1=z_2\), proving injectivity.
Any other holomorphic square root differs from \(g\) by a constant sign: on the connected punctured disk its quotient by \(g\) takes values in \(\{1,-1\}\) and is continuous, hence constant; continuity extends this equality across zero. Therefore either holomorphic branch has the asserted injectivity.