Problem 12 - Advanced Engineering Mathematics

Question

From the series \(e^{z}=-\sum_{k=0}^{\infty} \frac{(z-\pi i)^{k}}{k !}\) centered at \(\pi i\) and \\[ \begin{aligned} f(z) &=1-\pi i+z+e^{z}=1-\pi i+z+\left(-1-\frac{z-\pi i}{1 !}-\frac{(z-\pi i)^{2}}{2 !}-\frac{(z-\pi i)^{3}}{3 !}-\cdots\right) \\ &=-\frac{(z-\pi i)^{2}}{2 !}-\frac{(z-\pi i)^{3}}{3 !}-\cdots=(z-\pi i)^{2}\left(-\frac{1}{2 !}-\frac{z-\pi i}{3 !}-\cdots\right) \end{aligned} \\] we see that \(z=\pi i\) is a zero of order two

Step-by-Step Solution

Answer

The series indicates that \( z = \pi i \) is a zero of order two because the function \( f(z) \) starts with the term \( (z - \pi i)^2 \).

Step 1: Express Given Series

The exercise provides a function expansion for \( e^{z} \) centered at \( \pi i \), written as \( e^{z} = -\sum_{k=0}^{\infty} \frac{(z-\pi i)^{k}}{k !} \). This indicates that \( e^{z} \) is expressed as a power series centered at \( \pi i \).

Step 2: Understand the Function Definition

The function \( f(z) \) is given by \( f(z) = 1 - \pi i + z + e^{z} \). Substitute the series for \( e^{z} \) into this expression, which means\[ f(z) = 1 - \pi i + z + \left(-1-\frac{z-\pi i}{1 !}-\frac{(z-\pi i)^{2}}{2 !}-\cdots\right). \]

Step 3: Simplify the Function

By substituting and simplifying, we have:\[ f(z) = 1 - \pi i + z - 1 - \frac{z-\pi i}{1 !} - \frac{(z-\pi i)^{2}}{2 !} - \cdots = z - \pi i - \frac{z-\pi i}{1 !} - \frac{(z-\pi i)^{2}}{2 !} - \cdots. \] The constant term cancels out, focusing on terms with \( z-\pi i \).

Step 4: Reorganize and Factor Common Terms

Since the terms \((z-\pi i)^0\) and \((z-\pi i)^1\) do not appear after simplifying, the lowest power term is \( (z-\pi i)^{2} \). Hence,\[ f(z) = (z-\pi i)^{2} \left(-\frac{1}{2 !} - \frac{z-\pi i}{3 !} - \cdots\right). \]

Step 5: Conclude Zero Order

From the reorganization in Step 4, it is clear that \( z = \pi i \) is a zero of \( f(z) \) because the lowest power of \( (z-\pi i) \) is second order, confirming that \( \pi i \) is indeed a zero of order two.

Key Concepts

Power Series

Power series are an essential concept in complex analysis. They allow us to represent complex functions as infinite sums. Usually, power series have the form:\[ \sum_{k=0}^{\infty} a_k (z-c)^k \] where:

• \(a_k\) are the coefficients that define the series,
• \(z\) is the variable, representing a complex number,
• \(c\) is the center of the series, specifically a complex number around which the series is built.

This representation is powerful because it lets us understand the behavior of functions close to \(c\). It provides insight into the function's singularity and convergence.
The series becomes especially convenient when dealing with exponential functions, like the function \(e^z\) given in the exercise. Here, it's centered at \(\pi i\), transforming the exponential function into a power series. Power series also help decompose a function, making it easier to simplify and analyze. Recognizing these characteristics is crucial for solving problems related to complex functions.

Zeros of Functions

In complex analysis, finding the zeros of a function is a key task. A zero of a function \(f(z)\) is a point \(z_0\) where the function equals zero, meaning \(f(z_0) = 0\). Identifying zeros is important in understanding the function's properties and behavior.
They represent critical points where the function touches or crosses the x-axis in the complex plane.
Zeros can be described by their order, which indicates the multiplicity of a zero. If a function can be factored as \((z-z_0)^n g(z)\), where \(g(z)\) is non-zero at \(z_0\), then \(z_0\) is a zero of order \(n\). This concept is clearly illustrated in the exercise, where \(z = \pi i\) is shown to be a zero of order two.
Understanding the order provides critical insight into how the function behaves near its zeros.

Complex Functions

Complex functions extend the idea of functions beyond real numbers to the complex plane. These are functions where both the input and the output can be complex numbers. Such functions are central to the field of complex analysis.
They exhibit unique properties, such as being holomorphic or meromorphic, which refer to differentiability and the presence of poles, respectively.
In the exercise, the function \(f(z)\) involves the combination of linear terms and an exponential term expressed as a power series. This highlights the flexibility of complex functions. They can be constructed by combining simpler functions, each adding more layers to the function's behavior.
Complex functions are important because they offer rich insight and applications across mathematics and engineering. They allow us to solve advanced problems in fields like fluid dynamics, electromagnetism, and control systems. Understanding how to manipulate and decompose these functions is vital to unraveling the complexities of these advanced topics.