Let $D$ be a domain of the complex plane and the function $f\!:\,D \to \mathbb{C}$ be holomorphic. Then for each point $z$ of $D$ there is a corresponding point $w = f(z)\,\in \mathbb{C}$ ; we think that $z$ and $w$ both lie in their own complex planes, $z$ -plane and $w$ -plane.

Since $f$ is continuous in $D$ , if $z$ draws a continuous curve $\gamma$ in $D$ then its image point $w$ also draws a continuous curve $\gamma_w$ . Let $z_0$ and $z_0\!+\!\Delta z$ be two points on $\gamma$ and $w_0$ and $w_0\!+\!\Delta w$ their image points on $\gamma_w$ .

We suppose still that the curve $\gamma$ has a tangent line at the point $z_0$ and that the value of the derivative $f'$ has in $z_0$ a nonzero value

(1)

(2)

(3)

(4)

Theorem 1. If a curve $\gamma$ has a tangent line in a point $z_0$ where the derivative $f'$ does not vanish, then the image curve $f(\gamma)$ also has in the corresponding point $w_0$ a certain tangent line with a direction obtained by rotating the tangent of $\gamma$ by the angle $$\omega \;=\; \arg f'(z_0).$$

If the curve $\gamma$ is smooth, then also $\gamma_w$ is smooth, and it follows easily from (2) the corresponding limit equation between the arc lengths:

(5)

Conformality

If we have besides $\gamma$ another curve $\gamma'$ emanating from $z_0$ with its tangent, the mapping $f$ from $D$ in $z$ -plane to $w$ -plane gives two curves and their tangents emanating from $w_0$ . Thus we have two equations (4): $$\varphi_w \;=\; \varphi+\omega, \quad \varphi_w' \;=\; \varphi'+\omega$$ By subtracting we obtain

(6)

Theorem 2. The mapping created by the holomorphic function $f$ preserves the magnitude of the angle between two curves in any point $z$ where $f'(z) \neq 0$ . The equation (6) tells also that the orientation of the angle is preserved.

The facts in Theorem 2 are expressed so that the mapping is directly conformal. If the orientation were reversed the mapping were called inversely conformal; in this case $f$ were not holomorphic but antiholomorphic.