Conformality

A function $f:\mathbb{C}\to \mathbb{C}$ is conformal at $z_0$ if it preserves the angle (with orientation) between any two smooth curves intersecting at $z_0$.

If curves ${\gamma }_1,{\gamma }_2$ cross at $z_0$ at angle $\alpha$ (measured from ${\gamma }_1$ to ${\gamma }_2$ counterclockwise), their images $f({\gamma }_1),f({\gamma }_2)$ cross at $f(z_0)$ at the same angle $\alpha$ — direction and orientation both.

Theorem (analytic ⇒ conformal)

If $f$ is analytic at $z_0$ and $f^{\prime }(z_0)\ne 0$, then $f$ is conformal at $z_0$.

Why

Near $z_0$, by Taylor’s theorem, $f(z)\approx f(z_0)+f^{\prime }(z_0)(z-z_0)$. The map $z\mapsto z_0+f^{\prime }(z_0)(z-z_0)$ is multiplication by the complex constant $f^{\prime }(z_0)$. Multiplication by a complex number $w_0=|w_0|e^{i{\theta }_0}$ is a uniform rotation by ${\theta }_0$ and scaling by $|w_0|$ — applied uniformly to every direction.

Rotations preserve angles; uniform scaling preserves angles. So infinitesimally, $f$ acts like a rigid rotation-and-scaling, which is angle-preserving.

The condition $f^{\prime }(z_0)\ne 0$ is essential. If $f^{\prime }(z_0)=0$, the linear term vanishes. Expand $f$ around $z_0$ in a Taylor series:

$f(z)-f(z_0)=a_k(z-z_0{)}^k+a_{k+1}(z-z_0{)}^{k+1}+\cdots ,$

where $k\ge 2$ is the order of the lowest-order nonvanishing term (i.e., $f^{\prime }(z_0)=f^{\prime \prime }(z_0)=\cdots =f^{(k-1)}(z_0)=0$ but $f^{(k)}(z_0)\ne 0$). Near $z_0$, the dominant behavior is $f(z)-f(z_0)\approx a_k(z-z_0{)}^k$ — which raises angles to the $k$-th power, i.e., multiplies them by $k$.

The point $z_0$ is then a critical point of $f$, and conformality fails there.

Example: $f(z)=z^2$. At $z_0=0$, $f^{\prime }(0)=0$ and $f^{\prime \prime }(0)=2\ne 0$, so $k=2$. Curves through the origin have their crossing angles doubled under squaring.

Conjugation is anti-conformal

$f(z)=\bar{z}$ preserves the magnitude of angles but reverses orientation — a reflection. So conjugation is anti-conformal: angle-preserving up to a flip.

This is the cleanest geometric statement of why $\bar{z}$ is not analytic. Real-analytic-but-orientation-reversing maps fail the “rotation-and-scaling” character that complex differentiability demands.

Application: orthogonal coordinate systems

If $f=u+iv$ is analytic, the level curves of $u$ and the level curves of $v$ are mutually orthogonal.

Proof. $\nabla u=⟨u_x,u_y⟩$ and $\nabla v=⟨v_x,v_y⟩$. By Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$,

$\nabla u\cdot \nabla v=u_x{v}_x+u_y{v}_y=v_y{v}_x+(-v_x)v_y=0.$

Gradients are perpendicular, so the level curves they define are perpendicular.

Example. $f(z)=z^2$, so $u=x^2-y^2$ (level curves: hyperbolas with $x$- and $y$-axes as asymptotes) and $v=2xy$ (level curves: hyperbolas with $y=\pm x$ as asymptotes). These two families intersect at right angles — the curvilinear coordinate grid generated by squaring.

Why conformal maps matter in physics and engineering

Laplace’s equation is preserved by conformal maps. If $u$ is harmonic and $f$ is conformal, then $u\circ f^{-1}$ is harmonic in the image domain.

So you can solve Laplace problems by conformal transformation:

1. Find an analytic $f$ mapping your “hard” region (complicated boundary) to an “easy” region (disk, half-plane).
2. Solve Laplace’s equation with the transformed boundary conditions on the easy region.
3. Pull back to the original region via the inverse map.

This was the central technique of 19th-century electrostatics and fluid mechanics. The Smith chart in transmission-line theory is a working example: the Möbius map $\Gamma =(z-1)/(z+1)$ conformally maps the right half-plane to the unit disk, and properties of impedance in the right half-plane translate to properties of reflection coefficient on the chart.

The Riemann mapping theorem

The deepest result: any simply connected proper open subset of $\mathbb{C}$ can be mapped conformally onto the unit disk. So in principle, almost any 2D Laplace problem can be transformed to a disk problem. The practical issue is finding the explicit map — Möbius transformations and a few standard families handle the most useful cases.