Problem 5
Question
In Problems 1-6, use a Laurent series to find the indicated residue. $$ f(z)=e^{-2 / z^{2}} ; \operatorname{Res}(f(z), 0) $$
Step-by-Step Solution
Verified Answer
The residue of \( f(z) = e^{-2/z^2} \) at \( z = 0 \) is 0.
1Step 1: Recall Definition of a Laurent Series
A Laurent series is a representation of a complex function as a series which can have terms of negative degree. For a function \( f(z) \), around \( z_0 \), it is expressed as \( \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n \). The residue of \( f(z) \) at \( z_0 \) is the coefficient \( a_{-1} \).
2Step 2: Substitute and Expand the Exponential
Substitute the given function into the exponential series. Since \( f(z) = e^{-2/z^2} \), using the series expansion, we have \( e^{-2/z^2} = 1 - \frac{2}{z^2} + \frac{2^2}{2! z^4} - \frac{2^3}{3! z^6} + \ldots \) This is a series with only even negative powers.
3Step 3: Identify Terms of the Laurent Series
Since we need the residue, which is the coefficient of the \((z-0)^{-1}\) term, we look for terms in \( e^{-2/z^2} \) of the form \( \frac{1}{z} \). From the series expansion, all terms involve \(1/z^2\), \(1/z^4\), etc., leading to no \(1/z\) term.
4Step 4: Conclude on the Residue
Since there is no \( z^{-1} \) term in the expansion of \( e^{-2/z^2} \), the coefficient \( a_{-1} \) is 0. Consequently, the residue of \( f(z) \) at \( z = 0 \) is 0.
Key Concepts
Complex FunctionsSeries ExpansionResidue Calculation
Complex Functions
Complex functions are mathematical functions that involve complex numbers. A complex number is a number of the form \( z = x + yi \), where \( x \) and \( y \) are real numbers, and \( i \) is the imaginary unit with the property \( i^2 = -1 \). Complex functions extend the idea of functions from the real numbers to the complex numbers.
These functions can behave differently than real functions because they include oscillatory components. With their rich structure, complex functions have various applications, particularly in engineering and physics.
Understanding these functions is essential in complex analysis, where we study functions of a complex variable. In this context, tools like Laurent series (which we'll explore next) and residues become fundamental.
These functions can behave differently than real functions because they include oscillatory components. With their rich structure, complex functions have various applications, particularly in engineering and physics.
Understanding these functions is essential in complex analysis, where we study functions of a complex variable. In this context, tools like Laurent series (which we'll explore next) and residues become fundamental.
Series Expansion
Series expansion is a mathematical method of expressing a function as a sum of simpler terms. In the realm of complex functions, this can be particularly useful for understanding behavior near singularities or performing calculations.
One type of series expansion is the Laurent series, which is essential in complex analysis. It allows us to represent complex functions as a series, even in domains where Taylor series aren't applicable. The Laurent series for a function \( f(z) \) around a point \( z_0 \) is given by:
\[ f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n \]
This series includes terms with both positive and negative exponents, which helps in analyzing functions with singularities, such as poles.
When expanding a function like \( e^{-2/z^2} \), we substitute the complex function into a series form. This can help us find specific coefficients, like the residue, that give insights into the function’s properties near singular points.
One type of series expansion is the Laurent series, which is essential in complex analysis. It allows us to represent complex functions as a series, even in domains where Taylor series aren't applicable. The Laurent series for a function \( f(z) \) around a point \( z_0 \) is given by:
\[ f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n \]
This series includes terms with both positive and negative exponents, which helps in analyzing functions with singularities, such as poles.
When expanding a function like \( e^{-2/z^2} \), we substitute the complex function into a series form. This can help us find specific coefficients, like the residue, that give insights into the function’s properties near singular points.
Residue Calculation
Residue calculation is a powerful tool in complex analysis for evaluating certain types of integrals, especially in the presence of singularities. The residue of a function at a point \( z_0 \) is essentially the coefficient of \( (z-z_0)^{-1} \) in the Laurent series expansion.
To find the residue at \( z = 0 \) for a function like \( f(z) = e^{-2/z^2} \), we look at its series expansion. The goal is to identify the \( \frac{1}{z} \) term, as its coefficient is the residue.
For \( e^{-2/z^2} \), when expanded, it becomes clear that there are only terms with even negative powers: \( \frac{1}{z^2} \), \( \frac{1}{z^4} \), and so on. This means no \( \frac{1}{z} \) term is present. So, the residue at \( z = 0 \) is zero.
This process highlights the utility of residue calculations not just for integration, but also for understanding the nuances of complex functions around singular points.
To find the residue at \( z = 0 \) for a function like \( f(z) = e^{-2/z^2} \), we look at its series expansion. The goal is to identify the \( \frac{1}{z} \) term, as its coefficient is the residue.
For \( e^{-2/z^2} \), when expanded, it becomes clear that there are only terms with even negative powers: \( \frac{1}{z^2} \), \( \frac{1}{z^4} \), and so on. This means no \( \frac{1}{z} \) term is present. So, the residue at \( z = 0 \) is zero.
This process highlights the utility of residue calculations not just for integration, but also for understanding the nuances of complex functions around singular points.
Other exercises in this chapter
Problem 5
Determine the zeros and their orders for the given function. \(f(z)=z^{4}+z^{2}\)
View solution Problem 5
In Problems 1-10, evaluate the given trigonometric integral. $$ \int_{0}^{\pi} \frac{1}{2-\cos \theta} d \theta $$
View solution Problem 5
In Problems 3-8, determine the zeros and their orders for the given function. $$ f(z)=z^{4}+z^{2} $$
View solution Problem 5
In Problems 1-12, expand the given function in a Maclaurin series. Give the radius of convergence of each series. $$ f(z)=e^{-2 z} $$
View solution