Let \[ f:\Delta\to\mathbb C, \qquad f(z)=e^{z^2}. \] A disk \(B(p,r)\subset\mathbb C\) is called unramified for \(f\) if there is a holomorphic map \[ h:B(p,r)\to\Delta \] such that \(f(h(w))=w\) for every \(w\in B(p,r)\).
Find the radius of the largest unramified disk for \(f\).
Solution
The largest possible radius is \[ \boxed{\frac{e-1}{2}}. \]
1. Every unramified disk has radius at most \((e-1)/2\).
Write \[ f(z)=e^{z^2}. \] Since \(z^2\) maps \(\Delta\) onto \(\Delta\), the image of \(f\) is \[ \Omega=e^\Delta:=\{e^u:|u|<1\}. \] For \(u\in\Delta\), \[ |e^u-1| \le \sum_{n\ge1}\frac{|u|^n}{n!} =e^{|u|}-1 <e-1. \] Hence \[ \Omega\subset B(1,e-1). \]
Now suppose \(B(p,r)\) is unramified. Then \(B(p,r)\subset\Omega\), because every point of the disk is of the form \(f(h(w))\). Also \[ 1\notin B(p,r). \] Indeed, the only solution in \(\Delta\) of \(e^{z^2}=1\) is \(z=0\), and \(f'(0)=0\); differentiating \(f\circ h=\operatorname{id}\) at \(1\) would give the contradiction \[ 1=f'(0)h'(1)=0. \] Therefore \[ r\le |p-1|. \] On the other hand, the inclusion \[ B(p,r)\subset B(1,e-1) \] implies \[ |p-1|+r\le e-1. \] Combining the two inequalities gives \[ 2r\le e-1, \] so \[ r\le\frac{e-1}{2}. \]
2. Let\[ c=\frac{e+1}{2}, \qquad R=\frac{e-1}{2}. \]Then\[ B(c,R)\subset e^\Delta. \]
Consider the right half-disk \[ H=\{z:|z|<1,\ \operatorname{Re}z>0\}. \] The exponential is injective on \(\overline H\), so it maps \(H\) conformally onto a Jordan domain \(V=e^H\). We show that the boundary of \(V\) never enters \(B(c,R)\).
On the diameter of \(\partial H\), write \(z=iy\) with \(|y|\le1\). Then \[ \operatorname{Re}(e^{iy})=\cos y\le1=c-R, \] so \[ |e^{iy}-c|\ge R, \] with equality only at \(y=0\).
On the semicircular part of \(\partial H\), write \[ z=x+iy, \qquad x^2+y^2=1, \qquad 0\le x\le1. \] Since \[ c^2-R^2=e, \] we obtain \[ \begin{aligned} e^{-x}\bigl(|e^z-c|^2-R^2\bigr) &=e^x+e^{1-x}-(e+1)\cos y\\ &=:\Phi(x), \end{aligned} \] where \(|y|=\sqrt{1-x^2}\). We claim \(\Phi(x)\ge0\) on \([0,1]\).
Set \[ s=\sqrt{1-x^2}. \] Then \[ \Phi'(x) =e^x-e^{1-x}-(e+1)x\frac{\sin s}{s}, \] with \(\sin s/s=1\) at \(s=0\). If \(0\le x\le1/2\), the first difference is nonpositive, so \(\Phi'(x)<0\).
If \(1/2\le x\le1\), convexity of \(\sinh\) on \([0,1/2]\) gives \[ \sinh(x-1/2) \le2(x-1/2)\sinh(1/2). \] Thus \[ e^x-e^{1-x} \le4\sqrt e\,(x-1/2)\sinh(1/2). \] Since \(2(x-1/2)\le x\), \(\tanh(1/2)<1/2\), and \[ \frac{\sin s}{s}\ge1-\frac{s^2}{6}\ge\frac56, \] the last quantity is strictly smaller than \[ 2\sqrt e\,\cosh(1/2)\,x\frac{\sin s}{s} =(e+1)x\frac{\sin s}{s}. \] Hence \(\Phi'(x)<0\) throughout \([0,1]\). Since \[ \Phi(1)=e+1-(e+1)=0, \] we get \(\Phi(x)\ge0\), proving \[ |e^z-c|\ge R \] on the whole semicircle.
Therefore \[ \partial V\cap B(c,R)=\varnothing. \] But \(1<c<e\), so \(\log c\in(0,1)\) and hence \(c\in V\). If any point of the connected disk \(B(c,R)\) lay outside \(V\), the line segment from \(c\) to that point would meet \(\partial V\) inside \(B(c,R)\), a contradiction. Thus \[ B(c,R)\subset V\subset e^\Delta. \]
3. The disk \(B(c,R)\) is unramified for \(f\).
By 2, the principal logarithm maps \(B(c,R)\) holomorphically into \(\Delta\). Moreover \(1\) lies on the boundary of \(B(c,R)\), so \[ \log w\ne0 \qquad(w\in B(c,R)). \] Because the disk is simply connected, the zero-free holomorphic function \(\log w\) has a holomorphic square root there. Choose one branch and set \[ h(w)=\sqrt{\log w}. \] Then \[ |h(w)|^2=|\log w|<1, \] so \(h:B(c,R)\to\Delta\), and \[ f(h(w)) =e^{h(w)^2} =e^{\log w} =w. \] Thus \(B(c,R)\) is unramified. Together with 1, this proves that the largest radius is exactly \((e-1)/2\).