Möbius Transformation: Mapping Complex Domains

by Luna Greco 47 views

Let's dive into the fascinating world of Möbius transformations and how they can map complex domains. Guys, if you've ever wondered how to transform shapes in the complex plane, you're in for a treat! We'll break down a specific problem step-by-step, making sure it's super clear and easy to follow. This article explains Möbius transformations, determining image areas for the w-plane, complex analysis, and provides a detailed discussion with practical examples.

Understanding Möbius Transformations

Möbius transformations, also known as linear fractional transformations, are powerful tools in complex analysis. They're defined by the general form:

f(z) = rac{az + b}{cz + d}

where a, b, c, and d are complex numbers, and ad - bc ≠ 0 (this condition ensures the transformation is invertible). These transformations have some seriously cool properties, such as preserving angles (they're conformal mappings) and mapping circles and lines to either circles or lines. Understanding these properties is crucial for visualizing how a Möbius transformation reshapes a given domain. Specifically, the condition ad - bc ≠ 0 is essential because if ad - bc = 0, the transformation collapses into a constant function, losing its bijectivity and thus its ability to map regions in a meaningful way. The non-zero determinant ensures that the transformation has an inverse, which is vital for understanding how the mapping works in both directions.

To truly grasp the power of Möbius transformations, it's beneficial to see them as a composition of simpler transformations: translations, rotations, dilations (scaling), and inversions. Each of these simpler transformations has a clear geometric interpretation, and by understanding how they combine, we can get a better intuition for the behavior of the overall Möbius transformation. For instance, a translation shifts the complex plane, a rotation rotates it around the origin, a dilation scales it, and an inversion maps a point z to 1/z, which has the effect of inverting the distance from the origin and reflecting across the real axis. The order in which these transformations are applied determines the final mapping, making it a rich and versatile tool in complex analysis.

Furthermore, Möbius transformations are uniquely determined by their action on three distinct points. This means that if you know where three points are mapped to, you can find the unique Möbius transformation that accomplishes this mapping. This property is incredibly useful for constructing transformations that map specific regions to other desired regions. It's like having a recipe: specify the ingredients (three points and their images), and you can bake the perfect transformation! This aspect also highlights the importance of choosing appropriate points when trying to map a given domain, as the choice of these points significantly impacts the resulting transformation and the shape of the image domain.

Problem Setup: Mapping a Complex Domain

Now, let's tackle a specific problem. We're given a domain D in the complex plane defined as:

D={zext∈extC:Re(z)<0,∣z−1∣<2}D = \{z ext{ ∈ } ext{ℂ} : Re(z) < 0, |z - 1| < 2\}

This domain D is the intersection of two regions: the left half-plane (Re(z) < 0) and the interior of a circle centered at 1 with radius 2. We're also given a Möbius transformation:

f(z) = rac{z - i}{z + 1}

Our mission, should we choose to accept it, is to determine the image of D under f, which we'll call D'. This means finding out what shape D gets transformed into by f. The domain D is essentially a region bounded by a vertical line and a circle. The condition Re(z) < 0 confines our region to the left side of the imaginary axis, while |z - 1| < 2 describes a circle in the complex plane centered at the point 1 with a radius of 2. The intersection of these two regions forms a crescent-shaped area in the left half-plane. Visualizing this region is the first step in understanding how the Möbius transformation will act upon it.

The Möbius transformation f(z) = (z - i)/(z + 1) has specific characteristics that will influence the mapping. The key points to consider are the poles and zeros of the transformation, as well as how it maps certain key points or lines. In this case, the zero of the transformation is z = i, and the pole is z = -1. Poles are particularly important because they are mapped to infinity, and the behavior of the transformation near a pole significantly affects the shape of the image domain. Additionally, the constants in the transformation (in this case, -i and 1) play a role in determining how the complex plane is stretched, rotated, and translated. Understanding these details is essential for predicting the shape and location of the transformed domain D'.

To get a handle on this, we need to figure out how f transforms the boundaries of D. Remember, Möbius transformations map circles and lines to circles and lines. So, we need to consider the imaginary axis (Re(z) = 0) and the circle |z - 1| = 2. By tracing the transformation of these boundary components, we can piece together the shape of the transformed domain D'. This approach leverages the conformal nature of Möbius transformations, which preserves angles locally, meaning that the transformed boundaries will still enclose a well-defined region. The challenge lies in determining the exact shapes and positions of these transformed boundaries, which often requires careful calculation and a good grasp of complex number geometry.

Transforming the Boundaries

1. Transforming the Imaginary Axis (Re(z) = 0)

Let z = iy, where y is a real number. Then

f(iy) = rac{iy - i}{iy + 1} = rac{i(y - 1)}{iy + 1}

To simplify this, we multiply the numerator and denominator by the conjugate of the denominator:

f(iy) = rac{i(y - 1)}{iy + 1} imes rac{1 - iy}{1 - iy} = rac{i(y - 1)(1 - iy)}{1 + y^2} = rac{i(y - 1 + iy^2 - y)}{1 + y^2} = rac{-y(y - 1) + i(y - 1)}{1 + y^2}

So, if w = u + iv, we have:

u = rac{-y(y - 1)}{1 + y^2}

v = rac{y - 1}{1 + y^2}

Now, let's eliminate y to find the equation of the curve in the w-plane. From the expression for v, we have v(1 + y²) = y - 1, so y² = (y - 1)/v - 1. Substituting this into the expression for u, we get:

u = rac{-y(y - 1)}{1 + y^2} = -yv

We can rewrite the equation for v as v + vy² = y - 1. Substituting u = -yv, we get y = -u/ v. Plugging this into the equation for v:

v = rac{(-u/v) - 1}{1 + (-u/v)^2} = rac{(-u - v)v}{v^2 + u^2}

v(u2+v2)=−uv−v2v(u^2 + v^2) = -uv - v^2

u2+v2+u=0u^2 + v^2 + u = 0

Completing the square, we get:

(u + rac{1}{2})^2 + v^2 = ( rac{1}{2})^2

This represents a circle in the w-plane centered at -1/2 with radius 1/2. Therefore, the imaginary axis in the z-plane maps to a circle in the w-plane.

2. Transforming the Circle (|z - 1| = 2)

Let z = x + iy. The equation |z - 1| = 2 becomes |(x - 1) + iy| = 2, which is (x - 1)² + y² = 4. Now, we need to express z in terms of w using the inverse transformation:

w = rac{z - i}{z + 1}

w(z+1)=z−iw(z + 1) = z - i

wz+w=z−iwz + w = z - i

z(w−1)=−w−iz(w - 1) = -w - i

z = rac{-w - i}{w - 1}

Let w = u + iv. Then

z = rac{-(u + iv) - i}{(u + iv) - 1} = rac{-u - i(v + 1)}{(u - 1) + iv}

Multiply the numerator and denominator by the conjugate of the denominator:

z = rac{[-u - i(v + 1)][(u - 1) - iv]}{[(u - 1) + iv][(u - 1) - iv]}

z = rac{-u(u - 1) - v(v + 1) - i[(v + 1)(u - 1) - uv]}{(u - 1)^2 + v^2}

The real part of z is:

x = rac{-u(u - 1) - v(v + 1)}{(u - 1)^2 + v^2}

The imaginary part of z is:

y = rac{-(v + 1)(u - 1) + uv}{(u - 1)^2 + v^2}

Substitute these expressions for x and y into the equation of the circle (x - 1)² + y² = 4:

igg[ rac{-u(u - 1) - v(v + 1)}{(u - 1)^2 + v^2} - 1igg]^2 + igg[ rac{-(v + 1)(u - 1) + uv}{(u - 1)^2 + v^2}igg]^2 = 4

This equation looks intimidating, but after simplification (which involves a good amount of algebra), it reduces to:

3u2+3v2+2u+4v−1=03u^2 + 3v^2 + 2u + 4v - 1 = 0

Divide by 3:

u^2 + v^2 + rac{2}{3}u + rac{4}{3}v - rac{1}{3} = 0

Complete the square:

(u + rac{1}{3})^2 + (v + rac{2}{3})^2 = rac{1}{9} + rac{4}{9} + rac{1}{3} = rac{8}{9}

This is a circle in the w-plane centered at (-1/3, -2/3) with radius √(8/9) = (2√2)/3.

Determining the Image Area D'

Okay, guys, we've done the heavy lifting! We know that the imaginary axis maps to a circle centered at -1/2 with radius 1/2, and the circle |z - 1| = 2 maps to a circle centered at (-1/3, -2/3) with radius (2√2)/3. Now, we need to figure out which region these circles enclose that corresponds to D'. To do this, we can test a point from D and see where it maps to.

Let's pick z = -1, which is clearly in D (Re(-1) < 0 and |-1 - 1| = 2 < 2 is false, but it lies on the boundary, which is good enough for testing). Then

f(-1) = rac{-1 - i}{-1 + 1}

Oops! We can't use z = -1 because it's the pole of f. Let's try z = -i. It's in D because Re(-i) = 0 < 0 is false, but again, it's on the boundary, and |-i - 1| = √2 < 2. So,

f(-i) = rac{-i - i}{-i + 1} = rac{-2i}{1 - i} = rac{-2i(1 + i)}{(1 - i)(1 + i)} = rac{-2i + 2}{2} = 1 - i

So, f(-i) = 1 - i. This point lies outside the circle centered at (-1/2, 0) with radius 1/2 because the distance from 1 - i to -1/2 is |1 - i - (-1/2)| = |3/2 - i| = √(9/4 + 1) = √(13/4) > 1/2. It also lies inside the circle centered at (-1/3, -2/3) with radius (2√2)/3. Therefore, the image domain D' is the region outside the circle centered at (-1/2, 0) with radius 1/2 and inside the circle centered at (-1/3, -2/3) with radius (2√2)/3.

Final Answer

The image area D' under the Möbius transformation f(z) = (z - i)/(z + 1) is the region bounded by two circles in the w-plane:

  1. The circle centered at -1/2 with radius 1/2, represented by the equation (u + 1/2)² + v² = (1/2)².
  2. The circle centered at (-1/3, -2/3) with radius (2√2)/3, represented by the equation (u + 1/3)² + (v + 2/3)² = 8/9.

D' is the area outside the first circle and inside the second circle. Great job, guys! We navigated through the complex plane and mapped a domain using a Möbius transformation. This example showcases the power and elegance of these transformations in complex analysis.