The Riemann Mapping Theorem — Examples

The Cayley transform $C(z)=(z-i)\mathrm{/}(z+i)$ maps $\mathbb{H}$ biholomorphically onto $\mathbb{D}$, and multiplication by $i$ is a disc automorphism. Thus [L1] and [L2] make $f$ a biholomorphic map $\mathbb{H}\to \mathbb{D}$.

Direct substitution gives $f(i)=0$. Differentiating yields $f^{\mathrm{\prime }}(z)=i\frac{2i}{(z+i{)}^2},f^{\mathrm{\prime }}(i)=\frac{1}{2}>0.$ So $f$ has the required normalization at $i$.

The strip $S$ maps by $z\mapsto e^z$ biholomorphically onto the right half-plane, and the Cayley map $w\mapsto (w-1)\mathrm{/}(w+1)$ sends that half-plane biholomorphically onto $\mathbb{D}$. Hence $f$ is a biholomorphic map from $S$ onto $\mathbb{D}$.

Direct computation gives $f(0)=0$ and $f^{\mathrm{\prime }}(z)=\frac{2e^z}{(e^z+1{)}^2},f^{\mathrm{\prime }}(0)=\frac{1}{2}>0.$ So $f$ is normalized at the chosen basepoint.

By [L1], $z\mapsto z^2$ sends $S$ biholomorphically onto the right half-plane. Composing with the Cayley map $w\mapsto (w-1)\mathrm{/}(w+1)$ gives a biholomorphic map from $S$ onto $\mathbb{D}$, namely the displayed $f$.

Direct substitution gives $f(1)=0$, and $f^{\mathrm{\prime }}(z)=\frac{4z}{(z^2+1{)}^2},f^{\mathrm{\prime }}(1)=1>0.$ So the map is normalized at $1$.

By [L1], $\sqrt{z}$ maps $S$ biholomorphically onto the right half-plane. Composing with the Cayley map $w\mapsto (w-1)\mathrm{/}(w+1)$ gives the displayed biholomorphic map $f:S\to \mathbb{D}$.

One has $f(1)=0$, and $f^{\mathrm{\prime }}(z)=\frac{1}{\sqrt{z}(\sqrt{z}+1{)}^2},f^{\mathrm{\prime }}(1)=\frac{1}{4}>0.$ So $f$ is normalized at $1$.

Every map $h\in \mathcal{F}(\mathbb{D},0)$ satisfies $h(0)=0$ and $h^{\mathrm{\prime }}(0)>0$, so [L1] gives $0<h^{\mathrm{\prime }}(0)\le 1$.

The identity map belongs to $\mathcal{F}(\mathbb{D},0)$ and has derivative $1$ at $0$, so the extremal derivative is exactly $1$. Equality in [L1] forces any extremizer to be a rotation, and the positivity of the derivative leaves only the identity.

The function $k$ is holomorphic on $\mathbb{D}$, with $k(0)=0$ and $k^{\mathrm{\prime }}(0)=1$. The value $w=0$ has the unique preimage $z=0$. If $w\mathrm{\ne }0$, solving $w=z\mathrm{/}(1-z{)}^2$ gives $w{z}^2-(2w+1)z+w=0,$ so $z=\frac{2w+1\pm \sqrt{1+4w}}{2w}\mathrm{.}$ For $w\mathrm{\notin }(-\mathrm{\infty },-1\mathrm{/}4]$, choose the square-root branch that equals $1$ at $w=0$; then the minus sign gives the unique solution with $\mathrm{\mid }z\mathrm{\mid }<1$. Hence $k$ maps $\mathbb{D}$ bijectively onto $\mathbb{C}\setminus (-\mathrm{\infty },-1\mathrm{/}4]$.

The omitted point closest to $0$ is $-1\mathrm{/}4$, so no larger disc centered at $0$ can lie in $k(\mathbb{D})$. Since [L1] guarantees the quarter disc for every normalized univalent map, the constant $1\mathrm{/}4$ is sharp.

The inverse $g$ is bounded by $1$ on $\mathbb{D}\setminus \{0\}$, so [L1] extends it across the puncture to a holomorphic map $G:\mathbb{D}\to \mathbb{C}$. Continuity gives $\mathrm{\mid }G\mathrm{\mid }\le 1$ on the full disc.

The map $G$ is nonconstant because it agrees with the inverse $g$ off the puncture. If $\mathrm{\mid }G(0)\mathrm{\mid }=1$, [L2] would make $G(\mathbb{D})$ an open set containing a boundary point of the closed unit disc, contradicting $\mathrm{\mid }G\mathrm{\mid }\le 1$. Hence $G(0)\in \mathbb{D}$.

For every nonzero $z\in \mathbb{D}$, the inverse identity gives $f(G(z))=f(g(z))=z$. Letting $z\to 0$ and using step 2.1 plus continuity of $f$ gives $f(G(0))=0$. This contradicts $f(\mathbb{D})=\mathbb{D}\setminus \{0\}$, so no such biholomorphism exists.

By [L1], both $f(z)=z$ and $g(z)=-z$ are biholomorphic self-maps of $\mathbb{D}$.

The maps $f$ and $g$ are distinct, since $f(1\mathrm{/}2)=1\mathrm{/}2$ while $g(1\mathrm{/}2)=-1\mathrm{/}2$. Thus the disc already has two different conformal self-equivalences.

Therefore uniqueness fails unless a normalizing condition is imposed.

The map $f(z)=2z$ is holomorphic and bijects $D(0,1\mathrm{/}2)$ onto $\mathbb{D}$, with holomorphic inverse $w\mapsto w\mathrm{/}2$. Hence [L1] makes these two domains conformally equivalent.

Their Euclidean areas are different: $\mathrm{Area}(D(0,1\mathrm{/}2))=\frac{\pi }{4},\mathrm{Area}(\mathbb{D})=\pi \mathrm{.}$ So conformal equivalence does not preserve Euclidean area.