Math Problem Statement

Find where the Mobius transformation f(z)=1+z/1-z maps the unit circle , the real axis and the imaginary axis

Solution

To find where the Möbius transformation

f(z)=1+z1zf(z) = \frac{1 + z}{1 - z}

maps the unit circle, the real axis, and the imaginary axis, we need to analyze the image of each of these sets under the given transformation.

1. Mapping the Unit Circle

Let z=eiθz = e^{i\theta} represent points on the unit circle, where θ\theta ranges from 00 to 2π2\pi. Substituting z=eiθz = e^{i\theta} into the transformation, we get:

f(eiθ)=1+eiθ1eiθ.f(e^{i\theta}) = \frac{1 + e^{i\theta}}{1 - e^{i\theta}}.

Simplify this expression:

f(eiθ)=(1+eiθ)(1eiθ)(1eiθ)(1eiθ)=1eiθ+eiθ11eiθeiθ+1.f(e^{i\theta}) = \frac{(1 + e^{i\theta})(1 - e^{-i\theta})}{(1 - e^{i\theta})(1 - e^{-i\theta})} = \frac{1 - e^{-i\theta} + e^{i\theta} - 1}{1 - e^{i\theta} - e^{-i\theta} + 1}.

Simplify further:

f(eiθ)=eiθeiθ(eiθ+eiθ).f(e^{i\theta}) = \frac{e^{i\theta} - e^{-i\theta}}{-(e^{i\theta} + e^{-i\theta})}.

Using eiθeiθ=2isinθe^{i\theta} - e^{-i\theta} = 2i\sin\theta and eiθ+eiθ=2cosθe^{i\theta} + e^{-i\theta} = 2\cos\theta, the above simplifies to:

f(eiθ)=2isinθ2cosθ=itanθ.f(e^{i\theta}) = \frac{2i\sin\theta}{-2\cos\theta} = -i\tan\theta.

Thus, the unit circle maps to the imaginary axis.

2. Mapping the Real Axis

Let z=xz = x, where xx is a real number. Substituting z=xz = x into the transformation, we get:

f(x)=1+x1x.f(x) = \frac{1 + x}{1 - x}.

This is a real number as long as xx is real. Therefore, the real axis maps to the real axis but stretched according to the transformation.

3. Mapping the Imaginary Axis

Let z=iyz = iy, where yy is a real number. Substituting z=iyz = iy into the transformation, we have:

f(iy)=1+iy1iy.f(iy) = \frac{1 + iy}{1 - iy}.

Multiply numerator and denominator by the conjugate of the denominator:

f(iy)=(1+iy)(1+iy)(1iy)(1+iy)=1y2+2iy1+y2.f(iy) = \frac{(1 + iy)(1 + iy)}{(1 - iy)(1 + iy)} = \frac{1 - y^2 + 2iy}{1 + y^2}.

This can be rewritten as:

f(iy)=1y21+y2+i2y1+y2.f(iy) = \frac{1 - y^2}{1 + y^2} + i\frac{2y}{1 + y^2}.

This describes a transformation from the imaginary axis to the unit circle in the complex plane, parameterized by yy.

Summary

  • Unit Circle: Maps to the imaginary axis.
  • Real Axis: Maps to the real axis.
  • Imaginary Axis: Maps to the unit circle in the complex plane.

Would you like further details on any part of this transformation?

Here are some related questions that might help deepen your understanding:

  1. How can we find the fixed points of a Möbius transformation?
  2. What are the implications of a Möbius transformation preserving angles?
  3. Can Möbius transformations map a circle to a line? If so, how?
  4. How do Möbius transformations behave near their singularities?
  5. What are the properties of Möbius transformations in terms of conformal mappings?

Tip: Möbius transformations are powerful tools in complex analysis that preserve angles but not necessarily distances; understanding their geometric implications can greatly aid in visualizing complex mappings.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Möbius Transformation
Geometric Mappings

Formulas

Möbius transformation: f(z) = (1 + z)/(1 - z)
Euler's formula: z = e^(iθ) for points on the unit circle
Trigonometric identities: e^(iθ) - e^(-iθ) = 2i*sin(θ) and e^(iθ) + e^(-iθ) = 2*cos(θ)

Theorems

Properties of Möbius transformations (angle-preserving, mapping of lines and circles)

Suitable Grade Level

Undergraduate Level - Complex Analysis