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Chapter 2: Problem 11

Let \(0

Short Answer

Expert verified
Integrate using parameterization and cosine identities to evaluate both integrals.

Step by step solution

01

Rewrite the function

Let's start with the given function, \( f(z) = \frac{R+z}{(R-z)z} \). We want to express it in the form of \( f(z) = \frac{1}{z} + \frac{2}{R-z} \). First, rewrite the given function:\[f(z) = \frac{R+z}{(R-z)z} = \frac{R}{(R-z)z} + \frac{z}{(R-z)z}\]This simplifies to:\[= \frac{R}{(R-z)z} + \frac{1}{R-z}\]Thus, \[f(z) = \frac{1}{z} + \frac{2}{R-z} \].
02

Recognize and parameterize the curve \( \alpha \)

The curve \( \alpha \) is given by \( \alpha(t) = r \exp(\mathrm{i} t) \), with \( t \) ranging from 0 to \( 2\pi \). Using this parameterization, \( z = r\exp(\mathrm{i}t) \) traces out a circle of radius \( r \) centered at the origin in the complex plane.
03

Integrate over the curve \( \alpha \)

The integral to evaluate is:\[\frac{1}{2\pi} \int_0^{2\pi} \frac{R^2 - r^2}{R^2 - 2Rr \cos t + r^2} \, dt\]Notice \( z = r e^{it} \), let \( z = r e^{it} \) where \( \cos t = \frac{1}{2}(e^{it} + e^{-it}) \). Substitute all \( z \) using this relationship and simplify it to an expression that uses \( \cos t \). This integral evaluates to \( 1 \).
04

Showing the second integral

To prove:\[\frac{1}{2\pi} \int_0^{2\pi} \frac{R \cos t}{R^2 - 2Rr \cos t + r^2} \, dt = \frac{r}{R^2 - r^2}\]Use the same parameterization \( \cos t = \frac{1}{2}(e^{it} + e^{-it}) \) as before and substitute. Simplify this integral similar to the previous one and compute it to show the desired result.The key step involves stabilizing the form of the integral to cancel terms that don't alter the integration outcome. The use of symmetry and identities will assist in this task.

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Key Concepts

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

Contour Integration
Contour integration is a fundamental concept in complex analysis involving the integration of complex functions over certain paths or curves in the complex plane. These paths are known as contours. Contour integrals are a powerful tool when exploring the properties of complex functions, often revealing insights not obvious from the real analysis perspective. When calculating a contour integral, one must choose an appropriate path that is typically dictated by the problem at hand.

The key idea behind contour integration is to leverage the properties of analytic functions (functions that are complex differentiable) around a closed path. An important result is Cauchy鈥檚 integral theorem, which states that if a function is analytic within and on a closed contour C, then the integral of that function over C is zero. This provides a strong framework for dealing with integrals in the complex plane.
  • This type of integration can transform seemingly difficult real integrals into more manageable complex ones.
  • It often involves parametrizing a given curve or contour, which simplifies the integration process.
  • The selection of the contour can simplify, if not completely solve, various complex integrals using known theorems like the Cauchy integral theorem.
Parameterization of Curves
In complex analysis, parameterizing curves is a crucial step when carrying out contour integrations. By parameterizing a curve, we essentially describe it in a manner that makes it easier to apply integrals. For our exercise, the curve \(\alpha(t) = r \exp(i t)\) parameterizes a circle of radius \(r\) centered at the origin.

Each point along this curve is represented by substituting different values of \(t\), where \(t\) generally ranges over an interval, here from 0 to \(2\pi\). This approach allows us to transcribe a potentially complex path into a simpler, manageable set of equations that are much easier to interpret and work with during integration.
  • Parameterization helps convert complex variables to a linear time variable \(t\).
  • This facilitates ease of substitution into integrals, simplifying mathematics involved.
  • Using exponential forms, like \(\exp(it)\), ties in neatly with classic trigonometric identities, aiding in integral simplification.
Integral Evaluation
Integral evaluation is a core action in complex analysis, especially in verifying the results of contour integrals. The objective of integral evaluation is to solve or simplify integrals by appropriate substitution and algebraic manipulation.

In the exercise provided, we handle integrals such as \(\frac{1}{2\pi} \int_{0}^{2\pi} \frac{R^2 - r^2}{R^2 - 2Rr \cos t + r^2} \, dt\), where transformations and simplifications using trigonometric identities simplify the math. The cosine function, which appears due to the parameterization \(z = r e^{it}\), shows how integrating over the specified period results in specific expected outcomes, sometimes aided by symmetry considerations.
  • The symmetry and periodicity in trigonometric functions play a significant role in solving integrals.
  • Complementing parameterization helps in converting complex numbers back to a suitable real function form.
  • Integral identities such as trigonometric simplifications can be the key to solving or simplifying these evaluations.
Complex Functions
Complex functions are the bedrock of complex analysis, designed to take complex numbers as inputs and provide complex numbers as outputs. These functions exhibit properties like differentiability, which can greatly differ from real functions due to the presence of complex conjugates.

The function in our exercise \(f(z) = \frac{R+z}{(R-z)z}\) is an example, expressed further into simpler components like \(\frac{1}{z} + \frac{2}{R-z}\). This decomposition is useful for understanding the behavior and properties of the function across its domain. Analyzing these functions involves understanding their singularities, points where they do not behave as one might expect.
  • Complex functions often require rewriting or expansion to reveal simpler integral forms.
  • Handy identities in complex analysis facilitate the evaluation of functions when integrated.
  • Understanding these functions deeply can open paths to solving intricate integrals stemming from them.

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Most popular questions from this chapter

Continuous branches of the logarithm Let \(D \subset \mathbb{C}^{\bullet}\) be a domain which does not contain the origin. A continuous function \(l: D \rightarrow \mathbb{C}\) with \(\exp l(z)=z\) for all \(z \in D\) is called a continuous branch of the logarithm. Show: (a) Any other continuous branch of the logarithm \(\tilde{l}\) has the form \(\tilde{l}=\) \(l+2 \pi \mathrm{i} k, k \in \mathbb{Z}\) (b) Any continuous branch of the logarithm \(l\) is in fact analytic, and \(l^{\prime}(z)=1 / z\) (c) On \(D\) there exists a unique continuous branch of the logarithm only if the function \(1 / z\) has a primitive on \(D\). (d) Construct two domains \(D_{1}\) and \(D_{2}\) and continuous branches \(l_{1}: D_{1} \rightarrow\) \(\mathbb{C}, l_{2}: D_{2} \rightarrow \mathbb{C}\) of the logarithm, such that their difference is not constant on \(D_{1} \cap D_{2}\)

Determine all pairs \((f, g)\) of entire functions with the property $$ f^{2}+g^{2}=1 $$ Result: \(f=\cos \circ h\) and \(g=\sin \circ h\), where \(h\) is an arbitrary entire function.

Let \(D \subseteq \mathbb{C}\) be a domain and $$ f: D \longrightarrow \mathbb{C} $$ be an analytic function. Show: The function $$ \varphi: D \times D \longrightarrow \mathbb{C} $$ with $$ \varphi(\zeta, z):= \begin{cases}\frac{f(\zeta)-f(z)}{\zeta-z} & \text { if } \zeta \neq z \\ f^{\prime}(\zeta) & \text { if } \zeta=z\end{cases} $$ is a continuous function of two variables. For each given \(z \in D\) the function $$ \zeta \longmapsto \varphi(\zeta, z) $$ is analytic in \(D\).

Show that every rational function \(R\) (i.e. \(R(z)=P(z) / Q(z), P, Q\) polynomials, \(Q \neq 0)\) can be written as the sum of a polynomial and a finite linear combination, with complex coefficients, of "simple functions" of the form $$ z \mapsto \frac{1}{(z-s)^{n}}, \quad n \in \mathbb{N}, s \in \mathbb{C} $$ the so-called "partial fractions" (Partial fraction decomposition theorem), see also Chapter III, Appendix A to sections III.4 and III.5, Proposition A.7). Deduce: If the coefficients of \(P\) and \(Q\) are real, then \(f\) has "a real partial fraction decomposition" (by putting together pairs of complex conjugate zeros, or rather by putting together the corresponding partial fractions (see also Exercise 10 in I.1).

$$ \text { For } \alpha:[0,1] \rightarrow \mathbb{C} \text { with } \alpha(t)=\exp (2 \pi \mathrm{i} t) \text { compute } $$ $$ \int_{\alpha} 1 /|z| d z, \quad \int_{\alpha} 1 /\left(|z|^{2}\right) d z, \quad \text { and show } \quad\left|\int_{\alpha} 1 /(4+3 z) d z\right| \leq 2 \pi $$
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