Near a point where it is analytic, a function acts like multiplication by the number $f'(z_0)$: a turn and a uniform stretch. Mappings with that property keep the angles between curves, and that makes them the tool of choice for moving a hard problem (heat flow, fluid flow, electric fields) from an awkward region to a simple one.
Key idea
An analytic $f$ is conformal at $z_0$ when $f'(z_0) \ne 0$: any two smooth curves through $z_0$ that meet at an angle $\alpha$ have images that meet at the same angle $\alpha$, with the same orientation. Small shapes near $z_0$ are turned by $\arg f'(z_0)$ and scaled by $|f'(z_0)|$ (areas by $|f'(z_0)|^2$).
Where $f'(z_0) = 0$, angles are not kept: if the first nonzero derivative there is $f^{(n)}$, angles at $z_0$ are multiplied by $n$.
Where $f'(z_0) = 0$, angles are not kept: if the first nonzero derivative there is $f^{(n)}$, angles at $z_0$ are multiplied by $n$.
Why it works
A curve $z(t)$ through $z_0 = z(0)$ has tangent direction $z'(0)$. Its image $f(z(t))$ has tangent $f'(z_0)\,z'(0)$ by the chain rule. Multiplying by the same nonzero complex number $f'(z_0)$ turns every tangent by the same angle $\arg f'(z_0)$, so the angle between two tangents is unchanged. When $f'(z_0) = 0$ and $f(z) \approx f(z_0) + c(z - z_0)^n$ with $c \ne 0$, the map behaves like $(z - z_0)^n$, which multiplies angles at $z_0$ by $n$.
Worked examples
Worked example: Turn and stretch
For $f(z) = z^2$ at $z_0 = 1 + i$, find the angle by which small shapes are turned and the factor by which they are stretched.
- $f'(z_0) = 2(1 + i) = 2 + 2i = 2\sqrt{2}\,e^{\pi i/4}$.
- Turned by $\frac{\pi}{4}$, stretched by $2\sqrt{2}$ (areas by $8$).
Answer: angle $\frac{\pi}{4}$, factor $2\sqrt{2}$
Worked example: Where a map fails to be conformal
Where is $f(z) = z^3 - 3z$ not conformal?
- $f'(z) = 3z^2 - 3 = 3(z - 1)(z + 1)$.
- $f' = 0$ at $z = \pm 1$. There $f''(\pm 1) = \pm 6 \ne 0$, so angles at those points are doubled. Everywhere else $f$ is conformal.
Answer: at $z = \pm 1$
Worked example: Angles at a critical point
Two rays leave $0$ at an angle of $\frac{\pi}{6}$. At what angle do their images under $w = z^3$ leave $0$?
- A ray $\arg z = \theta$ goes to $\arg w = 3\theta$.
- So the angle becomes $3 \cdot \frac{\pi}{6} = \frac{\pi}{2}$.
Answer: $\frac{\pi}{2}$
Common mistake
Expecting angles to be kept where $f' = 0$. Conformality needs $f'(z_0) \ne 0$. The map $z^2$ is analytic everywhere, but at $0$ it doubles every angle: the positive real and positive imaginary axes (a right angle) go to the positive and negative real axes (a straight angle). Before claiming a map keeps angles at a point, compute $f'$ there.