Problem 17 - Complex Analysis

Question

Let us consider the upper half-plane \(\mathbb{H}=\\{z \in \mathbb{C} ; \quad \operatorname{Im} z>0\\}\) and the unit disk \(\mathbb{E}=\\{q \in \mathbb{C} ; \quad|q|<1\\}\) Show: The function $$ f(z):=\frac{z-\mathrm{i}}{z+\mathrm{i}} $$ provides a globally conformal map of H onto \(\mathbb{E}\). What is its inverse map? The map \(f\) is also called the Cayley map (A. CAYLEY, 1846 ).

Step-by-Step Solution

Answer

The Cayley map is conformal. Its inverse is \(z = i \cdot \frac{1+q}{1-q}\).

Step 1: Verify Conformality

To demonstrate that the function is a conformal map, we need to show that it is holomorphic and injective. The function \(f(z) = \frac{z-i}{z+i}\) is a ratio of two linear polynomial functions, which are holomorphic everywhere except where the denominator is zero. Thus, \(z\) cannot be equal to \(-i\). Since \(z\) is in the upper half-plane \(\mathbb{H}\), \(-i\) is not within the domain, confirming \(f\) is holomorphic on \(\mathbb{H}\).

Step 2: Check Injectivity

For \(f(z)\) to be injective, if \(f(z_1) = f(z_2)\), then \(z_1 = z_2\). Assume \(\frac{z_1-i}{z_1+i} = \frac{z_2-i}{z_2+i}\). Cross multiply to obtain \((z_1-i)(z_2+i) = (z_1+i)(z_2-i)\), which simplifies (by expanding and comparing real and imaginary parts) to \(z_1 = z_2\). Thus, \(f(z)\) is injective.

Step 3: Map \\( \\mathbb{H} \\rightarrow \\mathbb{E} \\)

Consider a point \(z\in\mathbb{H}\). Let \(z=x+iy\) where \(y>0\). We need to show \(|f(z)|<1\). Compute the modulus:\[|f(z)| = \left|\frac{z-i}{z+i}\right| = \frac{|z-i|}{|z+i|} = \frac{\sqrt{(x-0)^2+(y-1)^2}}{\sqrt{(x-0)^2+(y+1)^2}}\]Since \(y > 0\), the numerator \(y-1 < y+1\) of the denominator, hence \(|f(z)| < 1\), confirming \(f(z)\) maps \(\mathbb{H}\) into \(\mathbb{E}\).

Step 4: Derive Inverse Map

To find the inverse map, solve \(q = \frac{z-i}{z+i}\) for \(z\). Cross-multiply to get \(q(z+i) = z-i\) leading to \(qz + qi = z - i\). Rearranging terms gives:\[ (q-1)z = -i - qi \]Now solve for \(z\):\[z = \frac{-i - qi}{q-1} = i \cdot \frac{1+q}{1-q}\]Thus, the inverse map is \(z = i \cdot \frac{1+q}{1-q}\).

Key Concepts

Cayley Map

The Cayley map is named after the mathematician Arthur Cayley, known for his work in many areas of mathematics in the 19th century. Specifically, the Cayley map in our context serves as a bridge between two different regions in complex analysis: the upper half-plane and the unit disk. This remarkable function has a very specific form:

• It is given by the formula: \( f(z) = \frac{z-i}{z+i} \)
• Here, \(z\) is a complex number from the upper half-plane, meaning it must have a positive imaginary part.

The Cayley map transforms any point in the upper half-plane to a point within the unit disk. This means that every point on the line above the real axis gets mapped inside a circle in the complex plane. The boundary line where imaginary part equals zero gets mapped to the boundary of the unit circle. Thus, the entire upper half-plane is reshaped into the unit disk without any overlaps or gaps, maintaining conformality throughout the transformation.

Upper Half-Plane

The upper half-plane, denoted by \( \mathbb{H} \), refers to the set of complex numbers where the imaginary component is greater than zero.

• Formally, it's defined as \( \mathbb{H} = \{ z \in \mathbb{C} \mid \operatorname{Im} z > 0 \} \).
• In simpler terms, it includes all complex numbers that lie above the real axis in the complex plane.

This region is significant in various fields like complex analysis and differential equations.
In the problem at hand, using the mapping \( f \), we take complex numbers from this region and map them into another key geometric space, the unit disk. Understanding the upper half-plane and its properties is crucial for visualizing how these mappings operate and preserve the properties of the functions applied to them.

Holomorphic Function

A function is holomorphic if it is complex differentiable at every point in its domain. Holomorphic functions are a central concept in complex analysis because they are the complex-analog of differentiable functions in calculus.

• For the function \( f(z) = \frac{z-i}{z+i} \), it is holomorphic throughout the upper half-plane, except where the denominator becomes zero, which would occur only if \( z = -i \).

Since \( -i \) is not a part of the domain \( \mathbb{H} \) (because its imaginary part is not positive), \( f(z) \) remains holomorphic over the entire upper half-plane. This property ensures local angle-preserving transformations, reflecting the map's conformal nature as it morphs one geometric shape into another smoothly.
Holomorphicity is crucial when determining whether a function like our Cayley map can be termed as conformal, which allows it to preserve angles and the structure of small domains closely.

Injective Mapping

An injective mapping, or one-to-one function, ensures that each element of one set maps to a unique element of another set. Simply put, no two different points in the domain will have the same image in the co-domain.

• For our Cayley map \( f(z) = \frac{z-i}{z+i} \), the proof of injectivity involves showing that if \( f(z_1) = f(z_2) \), then it must follow that \( z_1 = z_2 \).
• The calculation in the solution, which involves setting the two expressions equal and simplifying, confirms the injectivity by resulting in \( z_1 = z_2 \).

Injectivity is a prerequisite for conformality because it guarantees that the mapping does not collapse distinct points onto the same image. Thus, in maintaining injectiveness, the function facilitates a true 1-to-1 correspondence between the upper half-plane \( \mathbb{H} \) and the unit disk \( \mathbb{E} \), making it possible to revert from \( f(z) \) to its inverse without ambiguity.