Möbius transformations and generalized circles

Try

The map T(z)=(z−i)/(z+i) sends i to zero and carries the upper half-plane into the unit disk. It turns a straight boundary line into a circle, with the missing boundary point corresponding to infinity.

A fractional-linear formula is not automatically a Möbius transformation. Its determinant ad−bc must be nonzero. Otherwise numerator and denominator can be proportional, giving a constant where defined rather than an invertible geometric map.

Predict

Predict where the pole of T lies. It is at −i, outside the upper half-plane, so the formula is holomorphic throughout that domain. Predict the image of a large positive real number: the value approaches one.

Infinity is not a large finite number. On the extended complex plane, or Riemann sphere, it is an additional point that makes these maps globally invertible and makes lines and circles part of one shared family.

Experiment

Track boundary and interior points under T(z)=(z−i)/(z+i).

Input	Output
i	0
0	−1
1	−i
Infinity	1

The real boundary passes through infinity on the sphere. Its image is the full unit circle, including the point one that no finite real input attains.

Understand

For T(z)=(az+b)/(cz+d) with ad−bc≠0, solving w=T(z) gives z=(dw−b)/(a−cw). If c≠0, the finite pole −d/c maps to infinity and infinity maps to a/c. If c=0, infinity maps to infinity. These conventions produce a bijection of the sphere. Away from the finite pole, T′(z)=(ad−bc)/(cz+d)² is nonzero.

To verify the half-plane map rather than relying on a diagram, write z=x+iy. For y>0, |z−i|²=x²+(y−1)² is smaller than |z+i|²=x²+(y+1)², so |T(z)|<1. The inverse is z=i(1+w)/(1−w). Its imaginary part equals (1−|w|²)/|1−w|², positive for |w|<1. This proves both inclusion and surjectivity.

Möbius maps carry generalized circles, meaning circles or lines together with the appropriate infinity point, to generalized circles. Such a set has equation A|z|²+Bz+B̄z̄+C=0, with A,C real. Substitute the inverse fractional-linear expression and clear the squared denominator. The resulting equation has the same form in w. On the sphere the map is bijective, so a genuine circle cannot collapse. A circle through the pole maps to a line, because its image contains infinity.

Images of three distinct points determine a unique Möbius map: mapping each triple to zero, one and infinity reduces uniqueness to a map fixing those three points.

Apply

Construct a map sending zero to zero, one to one and two to infinity. A zero at zero and pole at two suggest kz/(z−2); the value at one forces k=−1. Check T=z/(2−z) at all three points and verify its determinant is nonzero.

Diagnose “Every circle remains an ordinary finite circle.” A circle through the pole becomes a line. Ignoring infinity hides this case and can make a correct mapping look discontinuous merely because a finite chart has broken down.

Continue

The next lesson connects these exact maps with the local angle preservation of every holomorphic function having nonzero derivative. Optional depth: compose two fractional-linear formulas and compare their coefficient matrix with the product of their two-by-two matrices. Multiplying all four coefficients by one nonzero scalar leaves the same map.