Exercise 2105Algebra › Complex numbersM1Added 3 Nov 2026

The Möbius transformation

#Möbius#complex#transformation

Statement

Let $f(z)=\frac{az+b}{cz+d}$ with $ad-bc\neq0$ . Show that $f$ maps circles and lines to circles and lines.

Hint

The set of circles and lines in $\mathbb{C}$ is exactly the set of solutions of equations $\alpha|z|^2+\bar\beta z+\beta\bar z+\gamma=0$ .

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Solution

Equation of circles and lines. A generalized circle (circle or line) has equation $\alpha|z|^2+\bar\beta z+\beta\bar z+\gamma=0$ with $\alpha,\gamma\in\mathbb{R}$ and $(\alpha,\beta,\gamma)\neq(0,0,0)$ . Circles: $\alpha\neq0$ ; lines: $\alpha=0$ , $\beta\neq0$ .

Substitution. For $w=\frac{az+b}{cz+d}$ , $z=\frac{-dw+b}{cw-a}$ . Then substitute in the equation; the result is again of the form $\alpha'|w|^2+\bar\beta'w+\beta'\bar w+\gamma'=0$ .

Conclusion. Möbius transformations map generalized circles to generalized circles.

The Möbius transformation

Statement

Solution

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