Problem 31
Question
The function \(f(z)=\frac{1}{(z+2)(z-4 i)}\) possesses a Laurent series \(f(z)=\)
\(\sum_{k=-\infty}^{\infty} a_{k}(z+2)^{k}\) valid in the annulus \(r<|z+2|
Step-by-Step Solution
Verified Answer
The annulus has inner radius \( r = 0 \) and outer radius \( R = 2\sqrt{5} \).
1Step 1: Identifying Points of Singularity
The function \( f(z)=\frac{1}{(z+2)(z-4i)} \) is singular at points where the denominator is zero. Hence, the points of singularity are \( z = -2 \) and \( z = 4i \).
2Step 2: Determine the Center of Expansion
The Laurent series is expanded around \( z = -2 \). This means that the series will be expressed in terms of powers of \( (z+2) \).
3Step 3: Calculate Distances from Center to Singularities
Calculate the distance from the center of expansion \( z = -2 \) to each singularity. The distance to \( z = -2 \) itself is 0. To \( z=4i \), the distance is \(|-2 - 4i|=\sqrt{(-2)^2 + (4)^2} = \sqrt{20} = 2\sqrt{5}\).
4Step 4: Identify the Inner Radius \( r \)
Since the singularity at \( z = -2 \) is exactly at the center of expansion, \( r \) is the radius of the first non-zero distance from the center, which is 0 because any nearer singularities would be part of poles included in series. So \( r = 0 \).
5Step 5: Identify the Outer Radius \( R \)
The outer radius \( R \) is the smallest distance from the center of expansion to any singularity outside the immediate neighborhood considered by the given Laurent series. This is the distance to the next closest singularity, \( z = 4i \), which computed earlier to be \( 2\sqrt{5} \). Thus, \( R = 2\sqrt{5} \).
Key Concepts
Complex AnalysisSingularitiesRadius of ConvergenceAnnulus Domain
Complex Analysis
Complex analysis is a branch of mathematics that studies functions of complex variables. Unlike real numbers, complex numbers include an imaginary component, usually denoted by "i", where \(i^2 = -1\). This field is fundamental for understanding many mathematical and engineering concepts, including electrical circuits, fluid dynamics, and even quantum mechanics.
- Complex numbers can be represented in the form \(z = x + yi\), where \(x\) and \(y\) are real numbers.
- Functions of complex numbers can exhibit behaviors such as analyticity, meaning they can be represented by convergent power series.
Singularities
Singularities in complex analysis are points where a complex function becomes undefined or "blows up". These are crucial for determining the behavior of complex functions, as they can influence the convergence of series and the integrability of functions.
- Types of singularities include poles, essential singularities, and removable singularities.
- Pole: The function approaches infinity as it gets closer to the singularity. For the function \( f(z)=\frac{1}{(z+2)(z-4i)} \), \( z = -2 \) and \( z = 4i \) are poles.
- Essential Singularity: The function exhibits erratic behavior near the point.
- Removable Singularity: The function can be redefined so that it becomes analytic at that point.
Radius of Convergence
The radius of convergence is essential when dealing with power series in complex analysis. It is the radius within which a series converges to a function.
- For a power series \( \sum a_k (z-c)^k \), the center \(c\) is the point about which the series is expanded.
- The radius \( R \) can be calculated using the formula \( \frac{1}{R} = \limsup_{k \to \infty} \sqrt[k]{|a_k|} \), where \( a_k \) are the coefficients of the series.
Annulus Domain
An annulus domain in complex analysis is a ring-shaped region on the complex plane. This region is defined by two concentric circles centered at the same point but with different radii. For a series to converge in an annulus domain, it must converge between these circles.
- The inner radius \( r \) is the distance from the center to the closer circle.
- The outer radius \( R \) is the distance to the further circle.
Other exercises in this chapter
Problem 30
Find the circle and radius of convergence of the given power series. $$ \sum_{k=0}^{\infty} \frac{k !}{(2 k)^{k}} z^{3 k} $$
View solution Problem 31
Use Cauchy's residue theorem to evaluate the given integral along the indicated contour. $$ \oint_{C}\left(z^{2} e^{1 / \pi z}+\frac{z e^{z}}{z^{4}-\pi^{4}}\rig
View solution Problem 31
In Problems 31 and 32 , expand the given function in Taylor series centered at each of the indicated points. Give the radius of convergence \(R\) of each series
View solution Problem 31
Show that the power series \(\sum_{k=1}^{\infty} \frac{(z-i)^{k}}{k 2^{k}}\) is not absolutely convergent on its circle of convergence. Determine at least one p
View solution