📂Complex Anaylsis

Conformal Mapping Preserves the Angles

Theorem ¹

In the complex domain complex plane $\mathscr{R}$, let function $f$ be a conformal mapping, and let curves $\mathscr{C}_{1}$ and $\mathscr{C}_{2}$ meet at point $\alpha$ with an internal angle $\psi$.

If $\mathscr{C}_{1} '$ and $\mathscr{C}_{2} '$ are the images of $\mathscr{C}_{1}$ and $\mathscr{C}_{2}$ under $f$, respectively, then the two curves meet at $\beta = f ( \alpha )$ and their internal angle is also $\psi$.

Explanation

Though analytically worded in a challenging manner, the gist is that conformal mappings preserve the internal angles made by figures. The very name “conformal mapping” is derived from this property.

Meanwhile, a mapping that preserves the magnitude of angles but reverses their sign is called an Isogonal Mapping.

Proof

$f$ is a conformal mapping that sends $z = x + iy$ to $w = u + iv$.

Let the size of the internal angle formed by $\mathscr{C}_{1}$ and the $x$ axis be $\psi_{1}$, and let a point on $\mathscr{C}_{1}$ be $z_{1}$. Similarly, let the size of the internal angle formed by $\mathscr{C}_{2}$ and the $x$ axis be $\psi_{2}$, and let a point on $\mathscr{C}_{2}$ be $z_{2}$. Then, the internal angle formed by $\mathscr{C}_{1}$ and $\mathscr{C}_{2}$ would be $\psi_{2} - \psi_{1} = \psi$.

$$z - \alpha := r e^{i \theta_{1}} \\ z_{2} - \alpha = r e^{i \theta_{2}}$$ If we set it as $$\theta_{1} \to \psi_{1} \\ \theta_{2} \to \psi_{2}$$ when $r \to 0$ $$w_{1} - \beta = R_{1} e^{i \phi _{1}} \\ w_{2} - \beta = R_{2} e^{i \phi _{2}}$$ then. Assuming that $w_{k}: = f(z_{k})$ exists, we can set $f ' (\alpha) = \rho e^{ i \lambda }$ regarding $\rho > 0$.

$$f ’ ( \alpha) = \lim_{z_{1} \to \alpha } {{w_{1} - \beta } \over {z_{1} - \alpha }} = \lim_{z_{1} \to \alpha} {{R_{1}} \over {r}} e^{ i ( \phi_{1} - \theta_{1} )} = \rho e^{ i \lambda }$$ Thus, $$\lim_{z_{1} \to \alpha } (\phi_{1} - \theta_{1}) = \lambda$$ and accordingly, $$\lim_{w_{1} \to \beta } \phi_{1} = \psi_{1} + \lambda \\ \lim_{w_{2} \to \beta } \phi_{2} = \psi_{2} + \lambda$$ can be obtained. Therefore, the size of the internal angle formed by $\mathscr{C}_{1} '$ and the $u$ axis is $\psi_{1} + \lambda$ and the size of the internal angle formed by $\mathscr{C}_{2} '$ and the $u$ axis is $\psi_{2} + \lambda$. Finally, the internal angle formed by $\mathscr{C}_{1} '$ and $\mathscr{C}_{2} '$ cancels each other out as $\lambda$, $$(\psi_{2} + \lambda) - (\psi_{1} + \lambda) = \psi_{2} - \psi_{1} = \psi$$ .

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