Let $\textbf{x}(u,v)$ be a regular parametrized surface and let $z=u+iv$ denote the corresponding complex coordinate. Since $u=\frac{z+\bar{z}}{2}$ and $v=\frac{-i(z-\bar{z})}{2}$, we may write $\textbf{x}(z,\bar{z})=\left(x^1(z,\bar{z}),x^2(z,\bar{z}),x^3(z,\bar{z})\right).$

We define the complex function $\phi$ as follows,

$\phi(z):=\left(\frac{1}{2}\left(\frac{\partial x^1}{\partial u}-i\frac{\partial x^1}{\partial v}\right),\frac{1}{2}\left(\frac{\partial x^2}{\partial u}-i\frac{\partial x^2}{\partial v}\right),\frac{1}{2}\left(\frac{\partial x^3}{\partial u}-i\frac{\partial x^3}{\partial v}\right)\right)$

Theorem. Let $M$ be a surface with patch $\textbf{x}(u,v)$ and let $\phi=\frac{\partial\textbf{x}}{\partial z}$. Then $\textbf{x}(u,v)$ is isothermal if, and only if, $(\phi^1)^2+(\phi^2)^2+(\phi^3)^2=0$.

Proof. Suppose that $\textbf{x}(u,v)$ is isothermal. Note that $(\phi^i)^2=\frac{1}{4}[(x^i_u)^2-(x^i_v)^2-2ix^i_ux^i_v]$. Therefore,

$= (\phi)^2$

$= \frac{1}{4}\left[\sum_{i=1}^{3}(x^i_u)^2-\sum_{i=1}^{3}(x^i_v)^2-2i\sum_{i=1}^{3} x^i_ux^i_v\right]$

$= \frac{1}{4}\left(|\textbf{x}_u|^2-|\textbf{x}_v|^2-2i\langle\textbf{x}_u,\textbf{x}_v\rangle\right)$

$= \frac{1}{4}\left(E-G-2iF\right)$

$= 0$

Conversely, suppose that $(\phi)^2=0$. Then $\frac{1}{4}\left(E-G-2iF\right)=0$ and by properties of complex numbers this equations only holds if $E-G=0$ and $F=0$ which shows that $\textbf{x}(u,v)$ is isothermal. $\Box$

Theorem. Suppose that $M$ is a surface with patch $\textbf{x}(u,v)$. Let $\phi=\frac{\partial\textbf{x}}{\partial z}$ and suppose that $(\phi)^2=0$ (i.e. $\textbf{x}$ is isothermal). Then $M$ is minimal if, and only if, each $\phi^i$ is analytic.

Proof. Suppose that $M$ is minimal, then $\textbf{x}(u,v)$ is harmonic; that is, $\Delta\textbf{x}=0$. Thus, $\frac{\partial\phi}{\partial\bar{z}}=\frac{\partial}{\partial\bar{z}}\left(\frac{\partial\textbf{x}}{\partial z}\right)=\frac{1}{4}\Delta\textbf{x}=0$. Therefore, each $\phi^i=\frac{\partial\textbf{x}}{\partial z}$ is analytic. Conversely, the same calculation shows that if each $\phi^i$ is analytic, then each $x^i$ is harmonic, therefore, $M$ is minimal. $\Box$

Corollary. $x^i(z,\bar{z})=c_i+2\text{Re}\int\phi^i\;dz$.

Proof. Since $z=u+iv$, we may write $dz=du+idv$. Then

$\phi^idz=\frac{1}{2}\left[(x^i_u-ix^i_v)(du+idv)\right]=\frac{1}{2}\left[x^i_udu+x^i_vdv+i(x^i_udv-x^i_vdu)\right]$

$\bar{\phi}^idz=\frac{1}{2}\left[(x^i_u+ix^i_v)(du-idv)\right]=\frac{1}{2}\left[x^i_udu+x^i_vdv-i(x^i_udv-x^i_vdu)\right].$

We then have that $dx^i=\frac{\partial x^i}{\partial z}dz+\frac{\partial x^i}{\partial\bar{z}}d\bar{z}=\phi^idz+\bar{\phi}^id\bar{z}=2\text{Re}\phi dz$ and we can now integrate to get $x^i$. $\Box$

The problem of constructing minimal surfaces reduces to finding a tripple of analytic functions $\phi=(\phi^1,\phi^2,\phi^3)$ with $(\phi)^2=0$. A nice we of constructing such a $\phi$ is to take an analytic function $f$ and a meromorphic function $g$ with $fg^2$ analytic. Now, if we let $f=\phi^1-i\phi^2$ and $g=\frac{\phi^3}{(\phi^1-i\phi^2)}$ then we have,

$\phi^1=\frac{1}{2}f(1-g^2) \quad \phi^2=\frac{i}{2}f(1+g^2) \quad \phi^3=fg.$

Note that $f$ is analytic, $g$ is meromorphic, and $fg^2$ is analytic since $fg^2=-(\phi^1+i\phi^2)$. Furthermore, it is easily verified that $(\phi)^2=0$. Therefore, $\phi$ determines a minimal surface.

Theorem. (Weierstrass-Enneper Representation I) If $f$ is analytic on a domain $D,$ $g$ is meromorphic on $D$, and $fg^2$ is analytic on $D$ then a minimal surface is defined by $\textbf{x}(z,\bar{z})=\left(x^1(z,\bar{z}),x^2(z,\bar{z}),x^3(z,\bar{z})\right)$, where

$\displaystyle x^1(z,\bar{z})=\text{Re}\int f(1-g^2)\;dz$

$\displaystyle x^2(z,\bar{z})=\text{Re}\int if(1+g^2)\;dz$

$\displaystyle x^3(z,\bar{z})=\text{Re}2\int fg\;dz.$

Suppose that $g$ is analytic and has an inverse $g^{-1}$ in a domain $D$ which is analytic as well. Then we can consider $g$ as a new complex variable $\tau=g$ with $d\tau=g'dz$. Define $F(\tau)=\frac{f}{g'}$ and obtain $F(\tau)d\tau=fdz$. Therefore, if we replace $g$ by $\tau$ and $fdz$ by $F(\tau)d\tau$ we get the following.

Theorem. (Weierstrass-Enneper Representation II) For any analytic function $F(\tau)$, a minimal surface is defined by $\textbf{x}(z,\bar{z})=\left(x^1(z,\bar{z}),x^2(z,\bar{z}),x^3(z,\bar{z})\right)$, where

$\displaystyle x^1(z,\bar{z})=\text{Re}\int(1-\tau^2)F(\tau)\;d\tau$

$\displaystyle x^2(z,\bar{z})=\text{Re}\int i(1+g^2)F(\tau)\;d\tau$

$\displaystyle x^3(z,\bar{z})=\text{Re}2\int\tau F(\tau)\;d\tau.$

Note the corresponding $\phi=\left(\frac{1}{2}(1-\tau^2)F(\tau),\;\frac{i}{2}(1+\tau^2)F(\tau),\;\tau F(\tau)\right).$

This representaion tells us that any analytic function $F(\tau)$ defines a minimal surface.

We can now use the Weierstrass-Enneper representation to produce minimal surfaces. For example, if $(f, g) = (1, z)$ then we obtain a parametrization for Enneper’s surface. In fact, if $(f,g) = (1,z^n)$ then an nth order Enneper’s surface is obtained.

First order Enneper surface

The Weierstrass-Enneper representation leads to infinite families of minimal surfaces and has proved fundamental in relating the study of minimal surfaces to the theory of complex analysis.