Mobius Transformation

In complex analysis, a Mobius transformation (also known as a linear fractional transformation) is a special type of function that maps the extended complex plane (including the point at infinity) onto itself. These transformations play a fundamental role in conformal mapping, geometry, and various applications in mathematics and physics.

A Mobius transformation is defined as

w = f(z) = \frac{az + d}{cz + b}

where

• a, b, c, d are complex numbers such that ad − bc ≠ 0.

Domain and Range

• The transformation acts on the extended complex plane (also called the Riemann sphere), denoted by:

C∞ ​= C∪{∞}.

• It maps points in the complex plane and possibly ∞ as follows:

1. If z = −d/c, then the denominator is zero ⇒ w = ∞
2. If z = ∞, then w=a/c (when c ≠ 0)

Properties of Mobius Transformation

Some of the important properties are:

One-to-One and Onto Property:

Each point in the extended complex plane has a unique image under the transformation. Hence, the Mobius transformation is bijective on \mathbb{C}_\infty.

Preservation of Circles and Lines:

A Mobius transformation maps every circle or straight line in the z - plane to another circle or straight line in the w-plane.

Conformal (Angle-Preserving) Nature:

Mobius transformations are conformal mappings. They preserve the magnitude and orientation of angles between intersecting curves, except at points where the derivative f′(z) = 0.

Cross-Ratio Invariance:

The cross-ratio of four distinct points remains invariant under the transformation.

If z₁, z₂, z₃, z₄ map to w₁, w₂, w₃,w₄, then

\frac{(w_1 - w_3)(w_2 - w_4)}{(w_1 - w_4)(w_2 - w_3)} = \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)}.

This property is widely used in solving transformation problems.

Composition Property:

The composition of two Mobius transformations is again a Mobius transformation. Hence, the set of all Mobius transformations forms a group under composition.

Fixed Points:

A fixed point of a Mobius transformation is a point that remains unchanged under it, i.e., f(z) = z.

Solving, z = az + b/ cz + b, gives the quadratic equation:

cz²+ (d − a)z − b = 0.

Thus, a Mobius transformation can have two, one, or no fixed points.

Determinant Condition:

For the transformation to be valid and invertible, the determinant must satisfy: ad − bc ≠ 0.

Multiplying all coefficients by a common non-zero constant k does not change the transformation:

\frac{az + b}{cz + d} = \frac{k(az + b)}{k(cz + d)}.

Solved Question on Mobius Transformation

Question 1: Find the fixed point of f(z) = \frac{z + 2}{2z + 3}

Solution:

Set f(z) = z

z= \frac{z + 2}{2z + 3}

Cross-multiply

z(2z + 3) = z + 2
2z²+ 3z = z + 2
2z² + 3z − z − 2 = 0
2z² + 2z − 2 = 0

Divide by 2

z² + z − 1 = 0

Use quadratic formula

z = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}

Question 2: Consider the Möbius transformation f(z)=\frac{2z+1}{z+2}. Classify f as elliptic, parabolic, or hyperbolic (justify your classification).

Solution:

f(z)=\frac{2z+1}{z+2}.

f′(z) = \frac{2(z + 2)- (2z+1)(1)​ }{ (z + 2)^{2}}

= \frac{2z+4 - 2z -1}{(z+2)^2}

= \frac{3}{(z+2)^2}.

Evaluate at the fixed points:

• At z = 1: f'(1) = \frac{3}{(1 + 2)^{2}} = 3/9 = 1/3
• At z = -1: f'(-1) = \frac{3}{(-1+2)^2} = 3/1 = 3

So the multipliers at the two fixed points are 1/3 and 3. A Mobius transformation with two distinct fixed points and multipliers real and unequal to 1 is hyperbolic (it has an attractive fixed point and a repelling fixed point on the same invariant axis).

Question 3: Show that f(z) = 2z + 3\z + 1 is bijective on \widehat{\mathbb C}=\mathbb C\cup\{\infty\} and find its inverse.

Solution:

To find the inverse, set w = 2z + 3z + 1w =\dfrac{2z+3}{z+1} and solve for z:

w(z + 1) = 2z + 3 ⇒ wz + w = 2z + 3.

(w − 2)z = 3 − w ⇒ z = w − 2/3 − w​.

So f⁻¹(w) = 3 − w/w − 2​ for w ≠ 2, also check f( ⁣−1) = ∞ and f(∞) = 2 so the inverse handles ∞ correctly. Existence of this rational inverse shows f is bijective on \widehat{\mathbb C}.

Question 4: Verify f(z)=\dfrac{z-1}{z+1}​ that it is conformal at z = 2 by computing f′(2) and showing it is non-zero.

Solution:

Compute derivative:

f(z)=\dfrac{z-1}{z+1}

= \frac{(z+1)^2(1)(z+1) - (z-1)(1)}{(z+1)^2}

= \frac{(z+1) - (z-1)}{(z+1)^2}\

= \frac{z+1 - z + 1}{(z+1)^2}

= \frac{2}{(z+1)^2}.

Thus f′(2) = 2/9 ≠ 0. Since the derivative is nonzero, f is conformal (angle-preserving) at z = 2. (If f′(z₀) = 0 the map is not conformal at z₀)

Unsolved Questions on Mobius Transformation

Question 1: Find the fixed point(s) of f(z) = \frac{5z - 2}{2z + 3}

Question 2: Consider the Möbius transformation f(z) =\frac{4z-3}{z+5}.Determine whether the transformation is elliptic, parabolic, or hyperbolic (use the behavior of the derivative at the fixed points).

Question 3: Let g(z) = z+2/2z+3. Compute g′(z) and evaluate g′(1). Is g conformal at z = 1?

Question 4: Show that f(z) = 3z - 2\z + 4 is bijective on \widehat{\mathbb C}=\mathbb C\cup\{\infty\} and find its inverse.