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.
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If f is holomorphic at z_0 and
f'(z_0) \ne 0, then f is
conformal at z_0: it preserves the angle
between any two curves through z_0, in size and in sense.
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Where f'(z_0) = 0 the map is not conformal —
angles are multiplied. At such a point z^2 doubles
every angle, z^3 triples it, and so on.
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 tempting to shorten the slogan to "holomorphic maps are conformal" and forget
the fine print. The truth carries a crucial exception: a holomorphic f
is conformal only where f'(z_0) \ne 0. At a
critical point — a z_0 with
f'(z_0) = 0 — angles are not preserved; they are
multiplied. If f' vanishes to first order there
(so f''(z_0) \ne 0), the map behaves locally like
z^2 and doubles every angle, exactly as at the
origin above. Geometrically the plane gets folded or creased through such a point
rather than gently rotated — so "holomorphic ⟹ conformal" is only true away from the critical
points, and you must check f'(z_0) \ne 0 before you claim angles
survive.
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.