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For a path family $\Gamma$ on a Riemann surface, take the supremum of $L_{\rho }(\Gamma {)}^2/A_{\rho }$ over measurable conformal metrics of finite positive area. This is unchanged under conformal equivalence, and a quasiconformal map of maximal dilatation $K$ changes it by a factor between $K^{-1}$ and $K$.

A generalized conformal metric is locally $d{s}_{\rho }=\rho (z)|dz|$, where $\rho$ is nonnegative and measurable and transforms as a length density under a holomorphic change of coordinate. Its area is $A_{\rho }={\int }_X{\rho }^{2}dxdy$. Use the usual convention of using locally rectifiable paths for a path family $\Gamma$, and put $L_{\rho }(\Gamma )={\inf }_{\gamma \in \Gamma }{\int }_{\gamma }\rho |dz|$. ThenThis is extremal length. Zeros and isolated singularities of an admissible density are allowed; requiring a smooth strictly positive Riemannian metric would unnecessarily restrict the definition. Line integrals have their extended nonnegative values; if an arbitrary family is supplied, use its members that are locally rectifiable paths. An empty path family has infinite infimal length, whereas a family containing a constant path has extremal length zero.

Both numerator and denominator scale quadratically when $\rho$ is multiplied by a positive constant. The coordinate transformation of the area element makes the quotient unchanged under conformal equivalence.

The normalization relevant later is worth deriving. On the conformal cylinder $(\mathbb{R}/\mathbb{Z})\times (0,M)$, let $\Gamma$ contain the loops going once around it. For the horizontal loop at height $t$, Cauchy-Schwarz inequality givesIntegrating in $t$ shows $L_{\rho }(\Gamma {)}^2\le A_{\rho }/M$. The constant density achieves equality, since every winding-one loop has Euclidean length at least one. HenceHere $M$ is the height divided by circumference, the conformal modulus of an annulus; the reciprocal is used for the family joining its boundary components.

The boundary arcs meet at $0$ and $1$. The Möbius transformationsends these points to $0$ and infinity, sends the real boundary segment to the positive real axis, and sends the circular boundary arc to the ray of argument $-\pi /6$. Hence it maps $D\cap L$ onto the sectorThe power $\zeta =w^6$ maps this sector conformally onto the lower half-plane. Finally, the Cayley map $\zeta \mapsto (\zeta +i)/(\zeta -i)$ maps the lower half-plane onto the unit disc. Thus a requested conformal equivalence is

For the period lattice $\Lambda =⟨\lambda ,\mu ⟩$, define the Weierstrass elliptic function byTo prove Normal convergence of the Weierstrass elliptic-function series, let $K\subset \mathbb{C}\setminus \Lambda$ be compact and choose $R$ with $|z|\le R$ on $K$. For $|\omega |>2R$,The comparison series ${\sum }_{\omega \ne 0}|\omega {|}^{-3}$ converges by the stated lattice-sum criterion. The Weierstrass M-test gives uniform convergence on $K$; since $K$ was arbitrary, the series converges locally uniformly away from $\Lambda$. The locally uniform convergence of holomorphic functions makes its sum holomorphic there, and the displayed principal term gives a double pole at every lattice point.

The function $\wp$ is even and satisfiesThe three nonzero two-torsion points $z_1,z_2,z_3$ are distinct modulo $\Lambda$ and each equals its own negative modulo $\Lambda$. Therefore no two can have the same $\wp$-value, so $e_1,e_2,e_3$ are distinct. Equivalently, $\wp$ is a degree-two branched map and each half-lattice point already occupies its fiber with multiplicity two.

For a transitive example, take the equianharmonic latticeand define $\theta ([z])=[\omega z]$. Multiplication by $\omega$ preserves $\Lambda$, so $\theta :\mathbb{C}/\Lambda \to \mathbb{C}/\Lambda$ is a conformal equivalence. On the nonzero half-lattice points it acts bybecause ${\omega }^2=-1-\omega$. Thus it acts transitively.

Use the principal branch $\mathrm{Log}z=\log |z|+i\arg z$, with $-\pi <\arg z<\pi$. Thenis a conformal equivalence from the slit plane to the vertical strip. Its imaginary part tends to $-\infty$ as $z\to 0$ and to $+\infty$ as $z\to \infty$.

Every nonempty simply connected proper open subset of the complex plane is conformally equivalent to the unit disc.