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Chapter 2: Problem 15

Let \(D \subset \mathbb{C}\) be a domain with the property $$ z \in D \Rightarrow-z \in D $$ and \(f: D \rightarrow \mathbb{C}\) a continuous and even function \((f(z)=f(-z)) .\) Moreover, for some \(r>0\) let the closed disk \(\bar{U}_{r}(0)\) be contained in \(D .\) Then $$ \int_{a_{r}} f=0 \text { for } \alpha_{r}(t):=r \exp (2 \pi \mathrm{i} t), 0 \leq t \leq 1 $$

Short Answer

Expert verified
\(\int_{\alpha_r} f = 0\) due to symmetry and evenness of \(f\).

Step by step solution

01

Understand the Setup

We have a domain \( D \) in the complex plane such that if \( z \in D \), then \( -z \in D \). We are given a function \( f \) which is continuous and even, meaning \( f(z) = f(-z) \). Additionally, a closed disk centered at the origin with radius \( r \), denoted \( \overline{U}_r(0) \), is contained in \( D \).
02

Define the Path Integral

We need to evaluate the integral \( \int_{\alpha_r} f \), where \( \alpha_r(t) = r \exp(2 \pi i t) \) for \( 0 \leq t \leq 1 \). This path represents the boundary of the disk \( \overline{U}_r(0) \), which is a circle of radius \( r \) centered at the origin.
03

Analyze Symmetry and Evenness

Since \( f \'s \) evenness implies \( f(z) = f(-z) \) and since \( \, \alpha_r(t) = r \exp(2 \pi i t) \) runs over the circle, for each point \( z \) on the path there is a corresponding point \( -z \) due to the symmetry.
04

Consider the Integrals Over Symmetric Paths

The integral \( \int_{C} f(z) \, dz \) over a closed curve \( C \) which is symmetric around the origin can be split into two integrals over symmetric paths. For each pair of points \( z \) and \( -z \), the contributions to the integral cancel out because \( f(z) = f(-z) \) and the differential element is negated, \( dz = -d(-z) \).
05

Conclude the Integral is Zero

Due to the symmetry and the even nature of \( f \), the integral \( \int_{\alpha_r} f \) evaluates to zero as contributions over the symmetric paths cancel each other out. Therefore, the result is that the total integral over the circle is zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Even Function
An even function is a concept from mathematics where a function maintains symmetry with respect to the vertical axis. In the complex plane, this is expressed as \(f(z) = f(-z)\) for all \(z\) in its domain. This means that the value of the function at any point \(z\) is the same as its value at \(-z\).
  • Even functions are crucial when considering symmetrical situations, like in the domain given in the exercise.
  • The symmetry helps in simplifying calculations, especially when paired with definite integrals over symmetric paths.
In the context of complex numbers, this symmetry can result in cancellations when integrating over symmetric domains.
Path Integral
A path integral in complex analysis is an integral where the function is integrated over a path in the complex plane.
  • The path is represented as a parameterized curve \(\alpha(t)\) for \(t\) in some interval, often \([0, 1]\).
  • In the exercise, the path \(\alpha_r(t) = r \exp(2 \pi i t)\) traces out a circle of radius \(r\) centered at the origin.
Calculating the path integral involves evaluating how a function behaves as its input moves along this predefined path. The integral outcomes can be greatly influenced by the nature of both the path and the function itself.
Symmetric Domain
A symmetric domain in the context of complex analysis is a region in the complex plane that is symmetric with respect to some central point or axis. According to the exercise, if a point \(z\) is in the domain \(D\), then \(-z\) is also in \(D\).
  • This property introduces balance in analysis, which often leads to simplifications.
  • In conjunction with even functions, this symmetry can result in full cancellation during integration over the domain.
The symmetry can be leveraged to conclude that integrals over such domains might result in zero, thanks to the canceling pairs present due to the symmetric properties of both the domain and the function.
Complex Plane
The complex plane is a two-dimensional plane representing complex numbers, combining real and imaginary parts. Each complex number \(z = x + iy\) is represented as a point in this plane with coordinates \((x, y)\).
  • The horizontal axis represents the real part while the vertical axis represents the imaginary part.
  • Paths, domains, and symmetry related to complex functions can all be visualized within this plane.
Understanding the complex plane is essential for analyzing how complex functions behave and interact, including the visualization of domains like the ones in the exercise and how functions integrate along designated paths.
Continuous Function
A continuous function in mathematics is a function that does not have any abrupt changes, breaks, or gaps. Its output changes smoothly with changes in the input. In complex analysis, continuity ensures there are no sudden jumps in the values of the function when tracing paths in the complex plane.
  • This is essential when dealing with integrals over paths, as it guarantees the function can be smoothly integrated along any curve.
  • The exercise specifies that \(f\) is continuous to ensure the path integral is well-defined and manageable.
The combination of continuity with evenness of the function ensures that the calculations proceed smoothly and symmetrically, helping to reveal properties such as the integral resulting in zero.

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Most popular questions from this chapter

Which of the following subsets of \(\mathbb{C}\) are domains? (a) \(\left\\{z \in \mathbb{C} ; \quad\left|z^{2}-3\right|<1\right\\}\) (b) \(\left\\{z \in C ; \quad\left|z^{2}-1\right|<3\right\\}\) (c) \(\left\\{z \in \mathbb{C} ;\left.\quad|| z\right|^{2}-2 \mid<1\right\\}\) (d) \(\left\\{z \in C ; \quad\left|z^{2}-1\right|<1\right\\}\) (e) \(\\{z \in \mathbb{C} ; \quad z+|z| \neq 0\\}\) (f) \(\\{z \in \mathbb{C} ; 0

Let \(\omega\) and \(\omega^{\prime}\) be complex numbers which are linearly independent over \(\mathbb{R}\). Show: If \(f: \mathbb{C} \rightarrow \mathbb{C}\) is analytic and $$ f(z+\omega)=f(z)=f\left(z+\omega^{\prime}\right) \text { for all } z \in \mathbb{C} $$ then \(f\) is constant (J. LIOUVILLE, 1847 ).

Let \([a, b]\) and \([c, d]\) ( \(a

Let \(z_{0}, \ldots, z_{N} \in \mathbb{C}(N \in \mathbb{N}) .\) Define the line segments connecting \(z_{\nu}\) with \(z_{2+1}\) \((\nu=0,1, \ldots, N-1) \mathrm{by}\) $$ \alpha_{\nu}:[\nu, \nu+1] \longrightarrow \mathrm{C} \text { with } \alpha_{\nu}(t)=z_{\nu}+(t-\nu)\left(z_{\nu+1}-z_{\nu}\right) $$ Then \(\alpha:=\alpha_{1} \oplus \alpha_{2} \oplus \cdots \oplus \alpha_{N-1}\) defines a curve \(\alpha:[0, N] \rightarrow \mathbb{C} .\) One calls \(\alpha\) the polygonal path, which joins \(z_{0}\) with \(z_{N}\) (along \(\left.z_{1}, z_{2}, \ldots, z_{N-1}\right)\). Show: An open set \(D \subset \mathbb{C}\) is connected (and thus a domain) if and only if any two points of \(D\) can be connected by a polygonal path \(\alpha\) inside \(D\) (i.e. Image \(\alpha \subset D)\)

Let \(\alpha:[a, b] \rightarrow \mathbb{C}\) be continuously differentiable and assume that the function \(f:\) Image \(\alpha \rightarrow \mathbb{C}\) is continuous. Show: For any \(\varepsilon>0\) there exists a \(\delta>0\) with the following property: If \(\left\\{a_{0}, \ldots, a_{N}\right\\}\) and \(\left\\{c_{1}, \ldots, c_{N}\right\\}\) are finite subsets of \([a, b]\) with $$ a=a_{0} \leq c_{1} \leq a_{1} \leq c_{2} \leq a_{2} \leq \cdots \leq a_{N-1} \leq c_{N} \leq a_{N}=b $$ and $$ a_{\nu}-a_{\nu-1}<\delta \text { for } \nu=1, \ldots, N $$ then $$ \left|\int_{\alpha} f(z) d z-\sum_{\nu=1}^{N} f\left(\alpha\left(c_{\nu}\right)\right) \cdot\left(\alpha\left(a_{\nu}\right)-\alpha\left(a_{\nu-1}\right)\right)\right|<\epsilon $$ (Approximation of the line integral by a RIEMANN sum.)
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