Conformal Maps

A map of the plane can stretch, bend and fold. Some maps, though, are remarkably gentle: at a point they may rotate and rescale a tiny neighbourhood, but they leave every angle intact. Such a map is called conformal, and the surprise of this stage is that the holomorphic functions you already know are exactly the conformal ones.

Precisely: a map f is conformal at z_0 if, for any two smooth curves crossing at z_0, their images under f cross at f(z_0) at the same angle — same size and same sense (orientation). It preserves angles; it generally does not preserve lengths.

Why a nonzero derivative preserves angles

The whole proof is the local picture of the complex derivative together with the fact that complex multiplication is a rotate-and-scale.

Step 1 — linearise f near z_0. Differentiability means that for z close to z_0,

f(z) \approx f(z_0) + f'(z_0)\,(z - z_0).

So, up to the negligible higher-order part, the displacement z - z_0 is sent to the displacement f'(z_0)\,(z - z_0). Everything that happens to a tiny arrow leaving z_0 is just multiplication by the one fixed complex number f'(z_0).

Step 2 — read that multiplier in polar form. Write

f'(z_0) = |f'(z_0)|\,e^{i\varphi}, \qquad \varphi = \arg f'(z_0).

Multiplying any arrow by this number rotates it by the fixed angle \varphi and scales its length by the fixed factor |f'(z_0)|. This needs f'(z_0) \ne 0: if the multiplier were 0 there would be no direction to speak of.

Step 3 — apply it to two directions at once. Let two curves leave z_0 in directions making angles \alpha and \beta. After the map their tangent directions are

\alpha + \varphi \qquad \text{and} \qquad \beta + \varphi.

Step 4 — subtract. The same \varphi is added to both, so it cancels in the difference:

(\beta + \varphi) - (\alpha + \varphi) = \beta - \alpha.

The angle between the two curves is exactly what it was — and because both turned by the same \varphi in the same direction, the sense of the angle is preserved too. That is conformality.

Where it fails: f'(z_0) = 0

At a point with f'(z_0) = 0 the linear term vanishes and the next term takes over. For f(z) = z^2 at the origin, f'(0) = 0 and

f(z) = z^2 = \big(r e^{i\theta}\big)^2 = r^2 e^{i\,2\theta}.

A ray leaving the origin at angle \theta is sent to a ray at angle 2\theta, so the angle between two rays is doubled. Two perpendicular rays come out parallel. The map is angle-preserving everywhere except at the one point where its derivative dies.

It is worth savouring how little a conformal map is required to do. It may stretch a region enormously near one point and shrink it near another — the scale factor |f'(z)| roams freely. The single quantity it guards is the angle at which curves cross. That is why a fine square grid, pushed through a holomorphic map, comes out as a warped mesh whose every intersection is still a perfect right angle. This local rigidity is the engine behind everything in this stage: Möbius maps that send circles to circles, and the trick of solving a physics problem on an awkward shape by reshaping it conformally into an easy one.

Watch the crossing angle survive

Two arrows (blue) leave z_0 = e^{i\pi/4}; the dashed blue curves are the actual images of those directions under f(z) = z^2, and the orange arrows are their tangents at f(z_0). Here f'(z_0) = 2z_0 \ne 0, so the whole image bundle is rotated by \arg f'(z_0) = 45° and scaled — yet the angle between the two arrows is identical before and after. Slide the second direction \psi_2 and watch the two angle wedges stay equal.