Let’s understand the meaning of Mobius transformation in various contexts. Mobius transformation can be seen from a complex analysis and geometrical point of view. In geometry, we can get the Mobius transformation by performing multiple steps, such as
- Stereographic projection to the unit two-sphere from the plane,
- Rotating and moving the sphere to a new location,
- Orientation in space,
- Stereographic projection to the plane from the new position of the sphere.
Moreover, Mobius transformations preserve angles, map every circle to a line or circle and map every straight line to a line or circle.
Mobius Transformations in Complex Analysis
A complex function w = f(z) = (az + b)(cz + d); ad – bc ≠0 and a, b, c, d are complex numbers, and it is called a Mobius transformation or a fractional linear transformation.
The condition in Mobius transformations, i.e., ad – bc ≠0 ensures that the below statements hold.
- Neither az + b nor cz + d vanishes identically
- Both a and c cannot be equal to 0, where f would be constant
- Both b and d cannot be equal to 0, where f would be constant
- The denominator of f(z) cannot be a constant multiple of the numerator of this complex function.
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Properties of Mobius Transformations
A Mobius transformation is a composition of four elementary maps, namely translations, dilations, rotations, and inversions.
- Translations: z → z + z0 such that z0 ∈ C
- Dilations: z → λz; λ > 0 and λ ∈ R
- Rotations: z → eiθ z; θ ∈ R
- Inversions: z → 1/z
Fixed point of a Mobius Transformation
A point z0 ∈ C∞ is called a fixed point of the complex function f(z) if f (z0) = z0. A Mobius transformation can have at most two fixed points unless it is an identity map.
If z satisfies the condition f(z) = (az + b)/(cz + d) = z, then we can derive the following equations, i.e., cz2 – (a – d)z – b = 0.
When c = 0, z = -b/(a – d) is the only fixed point of f and which is ∞ if a = d.
Number of Mobius transformations
As we know, a Mobius transformation is completely determined by its action on three distinct points. Also, we can say that only one Mobius transformation is possible by its action on 3 distinct points in the complex plane C∞.
Cross-ratio
Suppose z1, z2, z3, z4 ∈ C∞ such that the cross-ratio of z1, z2, z3, z4 is a Mobius transformation defined by:
S(z) = (z1, z2, z3, z4) = (z – z3)(z2 – z4)/ (z2 – z3)(z – z4) such that S(z1) = 1, S(z3) = 0 and S(z4) = ∞.
When z2 = ∞, the cross-ratio (z1, z2, z3, z4) = (z1 – z3)/(z1 – z4)
When z3 = ∞, the cross-ratio (z1, z2, z3, z4) = (z2 – z4)/(z1 – z4)
When z4 = ∞, the cross-ratio (z1, z2, z3, z4) = (z1 – z3)/(z2 – z3)
Inverse of Mobius Transformations
Consider a Mobius transformation f(x) = (az + b)/(cz +d) is invertible from a complex plane to a complex plane. Then, the inverse of f(z), i.e., f-1(z) is again a Mobius transformation and is given as:
f-1(z) = (dz – b)/(-cz + a).
Mobius Transformations Solved Examples
Example 1:
Prove that a Mobius transformation takes circles into circles.
Solution:
Consider a complex function w = u + iv such that z = x + iy. From this, we can write the following:
u = x/(x2 + y2)
v = -y/(x2 + y2)
x = u/(u2 + v2)
y = -v/(u2 + v2)
We know that the Mobius transformation is the composition of translations, dilations and inversions.
Also, it is simple to show that translations and dilations take circles into circles.
So, let’s verify the inversion for circles into circles.
Consider the general equation of a circle:
A(x2 + y2) + BX + Cx + D = 0
Now, apply the transformation w = 1/z. So, substitute x = u/(u2 + v2) and y = -v/(u2 + v2) in the given circle equation.
Thus,
On simplification, we get;
⇒ A[1/ (u2 + v2) + B[u/(u2 + v2)] + C[-v/(u2 + v2)] + D = 0
⇒ [A + Bu + -Cv + D(u2 + v2)] /(u2 + v2) = 0
⇒ D(u2 + v2)] + Bu – Cv + A = 0
This is again a circle equation.
That means inversion preserves circles.
Hence proved.
Example 2:
Find the fixed point of f(z) = (3z – 1)/(z + 5).
Solution:
Given,
f(z) = (3z – 1)/(z +5)
Let f(z) = z
That means, (3z – 1)/(z + 5) = z
3z – 1 = z(z + 5)
3z – 1 = z2 + 5z
⇒ z2 + 5z – 3z + 1 = 0
⇒ z2 + 2z + 1 = 0
⇒ (z + 1)2 = 0
⇒ z = -1
Example 3:
Find a Mobius transformation with two fixed points, namely 3i and 1 + i.
Solution:
Given that 3i and 1 + i are the two fixed points of a Mobius transformation.
Let α and β be the two fixed points.
That means, α = 3i and β = 1 + i.
As we know, cz2 – (a – d)z – b = 0
This can be written as cz2 – (a – d)z – b = (z – α) (z – β)
From this, we can write the following:
c = 1
a – d = α + β = 3i + 1 + i = 1 + 4i
b = -αβ = -(3i)(1 + i) = -(3i + 3i2) = 3- 3i
Let k be any constant other than α or β.
Then one possible set of solutions with the given fixed points is:
a = k and d = k − α − β
We know that,
f(z) = (az + b)/(cz + d)
Now, by substituting the above values, we get;
f(z) = (kz – αβ)/ (z + k – α – β)
Let us assume that k = 1.
Therefore, f(z) = (z + 3 – 3i)/(z + 1 – 1 – 4i)
f(z) = (z + 3 – 3i)/(z – 4i)
Mobius Transformations Problems
- Find the fixed point of f(z) = 2z/(z + 1).
- Prove that every Mobius transformation is a composition of Translation, Dilation, Rotation and Inversion.
- Find the inverse of f(z) = (z + i)/(2z + 3i).
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