Problem 35 - A First Course in Complex Analysis with Applications

Question

By considering the series \(\sum_{k=0}^{\infty} r^{k} e^{i k \theta}, 0

Step-by-Step Solution

Answer

The given expressions are obtained by finding the real and imaginary parts of the geometric series sum.

Step 1: Understanding the Series

We start with the complex geometric series \(\sum_{k=0}^{\infty} r^k e^{i k \theta}\), where \(0 < r < 1\). This series represents an infinite sum of complex numbers, and it converges because the magnitude of the common ratio \(r\) is less than 1.

Step 2: Applying the Formula for Infinite Geometric Series

The sum of an infinite geometric series \(\sum_{k=0}^{\infty} x^k\) can be expressed as \(\frac{1}{1-x}\) for \(|x| < 1\). Exploiting this, we can write the sum for the complex series as \(\frac{1}{1-re^{i\theta}}\).

Step 3: Expressing the Denominator in an Alternate Form

The term \(1 - re^{i\theta}\) can be rewritten using exponential and trigonometric identities. We have \(e^{i\theta} = \cos \theta + i \sin \theta\), so \(re^{i\theta} = r\cos \theta + i r\sin \theta\). Thus, \(1 - re^{i\theta} = 1 - r\cos \theta - i r\sin \theta\).

Step 4: Finding the Magnitude Square of the Denominator

We need the magnitude of the denominator squared: \(|1 - re^{i\theta}|^2\). This equals \[ (1 - r\cos \theta)^2 + (r\sin \theta)^2 = 1 - 2r\cos \theta + r^2 \cos^2 \theta + r^2 \sin^2 \theta = 1 - 2r\cos \theta + r^2. \]This simplification works because \(\cos^2 \theta + \sin^2 \theta = 1\).

Step 5: Separating the Real and Imaginary Parts

The series \(\sum_{k=0}^{\infty} r^k e^{i k \theta}\) can be split into its real and imaginary components, i.e.,\[\sum_{k=0}^{\infty} r^k \cos k \theta + i \sum_{k=0}^{\infty} r^k \sin k \theta.\]The actual series, \(\frac{1}{1 - re^{i\theta}}\) gives this split when expressed as\[ \frac{1 - r\cos \theta}{1 - 2r\cos \theta + r^2} + i \frac{r\sin \theta}{1 - 2r\cos\theta + r^2}. \]

Step 6: Results - Real and Imaginary Sums

By equating the real parts and the imaginary parts from both the series form and the expression, we arrive at:1. For real parts: \(\sum_{k=0}^{\infty} r^k \cos k \theta = \frac{1 - r \cos \theta}{1 - 2r \cos \theta + r^2}\).2. For imaginary parts: \(\sum_{k=0}^{\infty} r^k \sin k \theta = \frac{r \sin \theta}{1 - 2r \cos \theta + r^2}\).

Key Concepts

Trigonometric Identities

When studying complex numbers and series, it's crucial to understand trigonometric identities, as these relationships help us transform and simplify expressions. One of the key identities is Euler's formula:

• \(e^{i\theta} = \cos \theta + i \sin \theta\)
• This identity bridges the gap between trigonometric functions and exponential functions and is often used to express complex numbers in polar form.

To break down the given series, we apply this identity to express complex numbers in terms of sines and cosines. By rewriting components like \(re^{i\theta}\) as \(r\cos \theta + ir\sin \theta\), it allows us to separate real and imaginary parts more clearly. Hence, trigonometric identities provide the mathematical tools to dissect complex series into more manageable parts.

Infinite Series Convergence

Infinite series can sometimes look daunting because they involve adding up an endless sequence of numbers. However, these series can converge, meaning they approach a specific value as more terms are added. The convergence criterion for the specific geometric series in this context is rooted in the series' common ratio, \(r\).

• If the absolute value of the common ratio \(|r| < 1\), then the series converges. In our problem, this is indeed true as \(0 < r < 1\).
• A convergent series has a finite sum, even if it extends to infinity.

Applying the convergence formula for a geometric series, \(\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}\) for \(|x| < 1\), allows us to express the sum of our complex series in a simplified form. This property makes infinite series a powerful tool in mathematical analysis.

Real and Imaginary Components

The decomposition of complex expressions into real and imaginary components is central in solving problems involving complex numbers. Given a complex number such as

• \(z = a + ib\), where \(a\) and \(b\) are real numbers, \(a\) represents the real component, and \(b\) represents the imaginary component.

In the case of our series, \(\sum_{k=0}^{\infty} r^k e^{i k \theta}\), we can separate it into

• \(\sum_{k=0}^{\infty} r^k \cos k \theta\) for the real part and
• \(\sum_{k=0}^{\infty} r^k \sin k \theta\) for the imaginary part.

This separation is essential for solving the exercise, as it allows us to equate the series' parts with their respective known expressions.

Complex Numbers

Complex numbers are integral to understanding the series in the exercise. They are numbers combined of both real and imaginary parts, commonly written in the form \(a + ib\). Each complex number can be represented using exponential functions, showcasing the interplay between trigonometry and complex numbers.

• The modulus or magnitude of a complex number \(z = a + ib\) is calculated as \(|z| = \sqrt{a^2 + b^2}\).
• This helps in finding the squared magnitude of expressions like \(\left| 1 - re^{i\theta} \right|^2 = 1 - 2r\cos \theta + r^2\), crucial for differentiating the real and imaginary sums.

Through the lens of these properties, we can easily dissect complex series into their corresponding components, making them easier to solve.