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Fix $z_0\in \mathcal{U}$. We prove the Riemann mapping theorem in this case by maximizing a normalized derivative. Choose $a\notin \mathcal{U}$. Since $\mathcal{U}$ is a simply connected domain and $z-a$ never vanishes, it has a holomorphic square root $g$. This $g$ is injective, and $g(\mathcal{U})$ is disjoint from $-g(\mathcal{U})$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal{U})$ and $r>0$ with $B(b,r)\subset g(\mathcal{U})$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded univalent function. Scaling and composing with an automorphism of the unit disk produces an injective holomorphic function $\varphi :\mathcal{U}\to \mathbb{D}$ with $\varphi (z_0)=0$ and ${\varphi }^{\prime }(z_0)>0$.

Let $\mathcal{F}$ be the family of all such normalized univalent functions. A Cauchy estimate in a small disk about $z_0$ bounds their derivatives, so $M={\sup }_{\varphi \in \mathcal{F}}{\varphi }^{\prime }(z_0)$ is finite and positive. Choose a maximizing sequence in this normal family. Montel theorem gives a subsequence converging uniformly on compact subsets to $\varphi$. Its derivative at $z_0$ is $M>0$. The maximum modulus principle puts its image in $\mathbb{D}$, and Hurwitz's theorem implies that a nonconstant limit of injective holomorphic functions is injective. Thus the maximum is attained.

If $a_1\in \mathbb{D}$ is omitted, then $a_1\ne 0$. Put $T_a(w)=(w-a)/(1-\overline{a}w)$. A holomorphic square root $v$ of $T_{a_1}\circ \varphi$ exists, is injective, and takes values in $\mathbb{D}$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in (0,1)$, the new derivative has magnitudea contradiction. Therefore $\varphi$ maps onto $\mathbb{D}$. The Cayley transform between the half-plane and disk now givesIts inverse is holomorphic by the holomorphic inverse function theorem, so this is a biholomorphism. The proper-subset hypothesis was used to choose $a\notin \mathcal{U}$; the whole complex plane cannot be mapped this way, by Liouville theorem.

On $\mathbb{C}\setminus [-1,1]$, the chosen branch of a multivalued function is characterized by $f(z{)}^2=z^2-1$ and $f(x)>0$ for $x>1$; equivalently $f(z)/z\to 1$ as $z\to \infty$. Define $h(z)=-f(-z)$. Then $h(z{)}^2=z^2-1$ and $h(z)/z\to 1$ at infinity, so uniqueness of the normalized holomorphic square root gives $h=f$. Therefore $f(-z)=-f(z)$.

On the slit domain $D$, choose the complex argument $\theta ={\mathrm{Arg}}_{(0,2\pi )}z$. The resulting holomorphic logarithm is $L(z)=\log |z|+i\theta$, with $L^{\prime }(z)=1/z$. DefineIt is holomorphic as a composition of holomorphic functions, and $f(z{)}^2=e^{L(z)}=z$. At $z=-1$ the chosen argument is $\pi$, so $f(-1)=i$.

Suppose a holomorphic square root $g$ existed on the whole punctured complex plane. It would be nonzero, and differentiating $g^2=z$ would give $2g{g}^{\prime }=1$. There are several ways to express the resulting obstruction; a direct one uses the unit circle. SetThe chain rule and $g^{\prime }(z)=g(z)/(2z)$ imply $q^{\prime }(t)=0$, so $q$ is constant. ButThese cannot agree because $g(1{)}^2=1$. Thus no such global holomorphic square root exists. Going once around zero changes the sign of the locally chosen root.

For a univalent function $f:\Omega \to \mathbb{D}$ with $f(z_0)=0$, suppose $a\ne 0$ is omitted. Apply the disk automorphism $T_a$ so the resulting image omits zero, take a holomorphic square root, then apply the disk automorphism taking the value at $z_0$ to zero. If $\Omega$ is a simply connected domain, the root exists. The resulting normalized univalent map has derivative magnitude multiplied by $(1+|a|)/(2\sqrt{|a|})>1$, since $(1-\sqrt{|a|}{)}^2>0$. Thus a map maximizing the derivative cannot omit an interior disk point.

The branch points are $\pm 1$; infinity has two unbranched meromorphic choices. On the complex plane slit along $[-1,1]$, normalization $f(z)\sim z$ at infinity selects a holomorphic square root with boundary values $f(x+i0)=i\sqrt{1-x^2}$ and $f(x-i0)=-i\sqrt{1-x^2}$ for $-1<x<1$. On the exterior real intervals, $f(x)=\mathrm{sgn}(x)\sqrt{x^2-1}$. A counterclockwise contour integral around the finite branch cut satisfies $\oint d\zeta /f(\zeta )=2\pi i$: the upper lip runs right to left, the lower lip left to right, and the endpoint arcs contribute $O(\sqrt{\epsilon })$. The opposite global branch reverses these signs.