ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13

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A Möbius map $(a z + b) / (c z + d)$ with $a d - b c \neq 0$ is conformal wherever $c z + d \neq 0$

Example

Let $a, b, c, d \in C$ with $a d - b c \neq 0$ , and define $F (z) = \frac{a z + b}{c z + d}$ wherever $c z + d \neq 0$ . Then $F$ is holomorphic and orientation-preserving conformal at every point of its domain.

Facts & Assumptions

Given: Complex coefficients $a, b, c, d$ with $a d - b c \neq 0$ .

[L1]

The linearity and quotient rules hold for complex derivatives wherever the denominator is nonzero (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L2]

A real-differentiable complex map is orientation-preserving conformal at a point exactly when it is complex differentiable there with nonzero derivative (A real-differentiable complex map is orientation-preserving conformal at a point exactly when it is complex differentiable there with nonzero derivative).

Verification

technique · direct computation

1.1

On the set $D : = {z : c z + d \neq 0}$ , [L1] gives $F^{'} (z) = \frac{a (c z + d) - c (a z + b)}{(c z + d)^{2}} = \frac{a d - b c}{(c z + d)^{2}} .$

L1algebra

2.1

Both numerator and denominator in step 1.1 are nonzero on $D$ , so $F^{'} (z) \neq 0$ throughout $D$ .

step 1.1given

3.1

Thus [L2] makes $F$ orientation-preserving conformal at every point of $D$ .

step 1.1step 2.1L2

4.1

If $c = 0$ , then $a d \neq 0$ , hence $d \neq 0$ and $D = C$ . If $c \neq 0$ , the excluded equation $c z + d = 0$ has the unique solution $z = - d / c$ , so $D = C ∖ {- d / c}$ . No global injectivity claim on $C$ is needed.

givenalgebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Howell and Mathews, Complex Analysis, §3.6 (standard reference, not scraped)

A Möbius map (az+b)/(cz+d)(az+b)/(cz+d) with ad−bc≠0ad−bc=0 is conformal wherever cz+d≠0cz+d=0