Conformal Mapping, Branches, and the Schwarz Lemma — Examples

Applying [F1] to $1=1(\cos 0+i\sin 0)$ gives $\mathrm{Log}(1)=0$.

Applying [F1] to $-1=1(\cos \pi +i\sin \pi )$ gives $\mathrm{Log}(-1)=i\pi$, hence $\mathrm{Log}(-1)+\mathrm{Log}(-1)=2\pi i\mathrm{\ne }0=\mathrm{Log}(1)=\mathrm{Log}((-1)(-1))$.

Since $1=1(\cos 0+i\sin 0)$, the principal square root of $1$ is $1$.

By [F1], $\text{exception 25:}=e^{i\pi \mathrm{/}2}=i$, hence $\text{exception 25:}\text{exception 25:}=i^2=-1\mathrm{\ne }1=\text{exception 25:}=\text{exception 25:}$.

If $z=x+iy\in S$, then $0<y<\pi$, so $e^z=e^x(\cos y+i\sin y)$ has imaginary part $e^x\sin y>0$; hence $e^z\in \mathbb{H}$.

For $w=u+iv\in \mathbb{H}$ one has $\mathrm{\mid }w-i{\mathrm{\mid }}^2=u^2+(v-1{)}^2<u^2+(v+1{)}^2=\mathrm{\mid }w+i{\mathrm{\mid }}^2$, so $C(w):=(w-i)\mathrm{/}(w+i)$ satisfies $\mathrm{\mid }C(w)\mathrm{\mid }<1$ and maps $\mathbb{H}$ into $\mathbb{D}$.

The inverse Möbius map is $C^{-1}(\zeta )=i(1+\zeta )\mathrm{/}(1-\zeta )$; for $\mathrm{\mid }\zeta \mathrm{\mid }<1$, the identity $\mathrm{Im}\text{\hspace{0.17em}⁣}\left(i\frac{1+\zeta }{1-\zeta }\right)=\frac{1-\mathrm{\mid }\zeta {\mathrm{\mid }}^2}{\mathrm{\mid }1-\zeta {\mathrm{\mid }}^2}>0$ shows $C^{-1}(\zeta )\in \mathbb{H}$, so [F2] confirms that this Cayley map is exactly the standard upper-half-plane automorphism sending $\mathbb{H}$ biholomorphically to $\mathbb{D}$. Together with step 1.1, this makes $\mathrm{\Phi }=C\circ e^z:S\to \mathbb{D}$ biholomorphic.

By [F1], both ${\phi }_a$ and ${\phi }_b$ are automorphisms of $\mathbb{D}$, so their composition $f_{a,b}={\phi }_b\circ {\phi }_a$ is again an automorphism of $\mathbb{D}$.

Since ${\phi }_a(a)=0$ and ${\phi }_b(0)=b$, one has $f_{a,b}(a)={\phi }_b({\phi }_a(a))={\phi }_b(0)=b$.

The argument interval of $S$ is $(-\pi \mathrm{/}4,\pi \mathrm{/}4)$, which has width $\pi \mathrm{/}2<\pi$, so [F1] applies to $z\mapsto z^2$.

Doubling the argument interval gives $(-\pi \mathrm{/}2,\pi \mathrm{/}2)$, hence the image is $\{\rho e^{i\varphi }:\rho >0,-\pi \mathrm{/}2<\varphi <\pi \mathrm{/}2\}$, which is exactly the right half-plane. Therefore $z\mapsto z^2$ is a biholomorphism from $S$ onto $\{\mathrm{Re}w>0\}$.

Parameterizing the circle by $z=r{e}^{it}$, $0\le t<2\pi$, [F1] gives $J(r{e}^{it})=\frac{1}{2}(r{e}^{it}+r^{-1}{e}^{-it})=\frac{r+r^{-1}}{2}\cos t+i\frac{r-r^{-1}}{2}\sin t$.

Writing $x=\frac{r+r^{-1}}{2}\cos t$ and $y=\frac{r-r^{-1}}{2}\sin t$, one gets $\frac{x^2}{{\left(\frac{r+r^{-1}}{2}\right)}^2}+\frac{y^2}{{\left(\frac{r-r^{-1}}{2}\right)}^2}={\cos }^{2}t+{\sin }^{2}t=1$, so the image is the stated ellipse.

For $y>0$, $\sin (\pi \mathrm{/}2+iy)=\sin (\pi \mathrm{/}2)\cosh y+i\cos (\pi \mathrm{/}2)\sinh y=\cosh y>0$, and similarly $\sin (-\pi \mathrm{/}2+iy)=-\cosh y<0$.

For real $x\in (-\pi \mathrm{/}2,\pi \mathrm{/}2)$ one has $\sin x\in (-1,1)\subset \mathbb{R}$; together with step 1.1, every boundary component of $S$ maps into $\mathbb{R}$, and [F1] identifies the interior image as $\mathbb{H}$.

On tangent vectors at $0$, $c$ sends $1\mapsto 1$ and $i\mapsto -i$, so the unoriented angle still has magnitude $\pi \mathrm{/}2$ but the oriented angle changes from $+\pi \mathrm{/}2$ to $-\pi \mathrm{/}2$.

The complex difference quotient at $0$ is $(c(h)-c(0))\mathrm{/}h=\bar{h}\mathrm{/}h$; along real $h\to 0$ this equals $1$, while along purely imaginary $h\to 0$ it equals $-1$, so the limit does not exist, $c$ is not holomorphic, and [F1] therefore excludes it from being conformal in the library's sense.

The map $f(z)=2z$ is holomorphic on $\mathbb{C}$, bijective, and has holomorphic inverse $f^{-1}(w)=w\mathrm{/}2$, so [F2] and [F1] make it conformal.

But the unit tangent vector $1$ at $0$ is sent to $f^{\mathrm{\prime }}(0)\cdot 1=2$, whose Euclidean length is $2\mathrm{\ne }1$. Therefore a conformal map need not preserve Euclidean lengths.

By [F1], $f(z)={\phi }_{1\mathrm{/}2}(z)$ is a holomorphic self-map of $\mathbb{D}$.

But $f(0)=1\mathrm{/}2$, so $\mathrm{\mid }f(0)\mathrm{\mid }=1\mathrm{/}2>0=\mathrm{\mid }0\mathrm{\mid }$; this already violates the usual Schwarz-lemma bound $\mathrm{\mid }f(z)\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }$ at $z=0$. Hence the fixed-point hypothesis at $0$ cannot be removed.