11.3: Analytic functions are Conformal

Theorem $\PageIndex{1}$ Operational definition of conformal

If $f$ is analytic on the region $A$ and $f'(z_0) \ne 0$, then $f$ is conformal at $z_0$. Furthermore, the map $f$ multiplies tangent vectors at $z_0$ by $f'(z_0)$.

Proof

The proof is a quick computation. Suppose $z = \gamma (t)$ is curve through $z_0$ with $\gamma (t_0) = z_0$. The curve $\gamma (t)$ is transformed by $f$ to the curve $w = f(\gamma (t))$. By the chain rule we have

$$\dfrac{d f(\gamma (t))}{dt} \vert_{t_0} = f'(\gamma (t_0)) \gamma ' (t_0) = f'(z_0) \gamma ' (t_0). \nonumber$$

The theorem now follows from Theorem 11.3.1.

Example $\PageIndex{1}$ Basic example

Suppose $c = ae^{i \phi}$ and consider the map $f(z) = cz$. Geometrically, this map rotates every point by $\phi$ and scales it by $a$. Therefore, it must have the same effect on all tangent vectors to curves. Indeed, $f$ is analytic and $f'(z) = c$ is constant.

Example $\PageIndex{2}$

Let $f(z) = z^2$. So $f'(z) = 2z$. Thus the map $f$ has a different affect on tangent vectors at different points $z_1$ and $z_2$.

Example $\PageIndex{3}$ Linear approximation

Suppose $f(z)$ is analytic at $z = 0$. The linear approximation (first two terms of the Taylor series) is

$$f(z) \approx f(0) + f'(0) z. \nonumber$$

If $\gamma (t)$ is a curve with $\gamma (t_0) = 0$ then, near $t_0$,

$$f(\gamma (t)) \approx f(0) + f'(0) \gamma (t). \nonumber$$

That is, near 0, $f$ looks like our basic example plus a shift by $f(0)$.

Example $\PageIndex{4}$

The map $f(z) = \overline{z}$ has lots of nice geometric properties, but it is not conformal. It preserves the length of tangent vectors and the angle between tangent vectors. The reason it isn’t conformal is that is does not rotate tangent vectors. Instead, it reflects them across the $x$-axis.

In other words, it reverses the orientation of a pair of vectors. Our definition of conformal maps requires that it preserves orientation.