Problem 11
Question
$$\begin{array}{l} \operatorname{Res}(f(z),-1)=\lim _{z \rightarrow-1}(z+1) \cdot \frac{5 z^{2}-4 z+3}{(z+1)(z+2)(z+3)}=\lim _{z \rightarrow-1} \frac{5 z^{2}-4 z+3}{(z+2)(z+3)}=6 \\ \operatorname{Res}(f(z),-2)=\lim _{z \rightarrow-2}(z+2) \cdot \frac{5 z^{2}-4 z+3}{(z+1)(z+2)(z+3)}=\lim _{z \rightarrow-2} \frac{5 z^{2}-4 z+3}{(z+1)(z+3)}=-31 \\ \operatorname{Res}(f(z),-3)=\lim _{z \rightarrow-3}(z+3) \cdot \frac{5 z^{2}-4 z+3}{(z+1)(z+2)(z+3)}=\lim _{z \rightarrow-3} \frac{5 z^{2}-4 z+3}{(z+1)(z+2)}=30 \end{array}$$
Step-by-Step Solution
Verified Answer
The residues are 6 at \(-1\), -31 at \(-2\), and 30 at \(-3\).
1Step 1: Understanding the Residue Formula
To find the residue of a function at a simple pole, we use the formula \( \operatorname{Res}(f(z), a) = \lim_{z \rightarrow a} (z-a) \cdot f(z) \). Here, \( a \) is the pole of the function.
2Step 2: Find \( \operatorname{Res}(f(z), -1) \)
Identify that \( z = -1 \) is a pole of \( f(z) = \frac{5z^2 - 4z + 3}{(z+1)(z+2)(z+3)} \). The residue is calculated as: \[ \operatorname{Res}(f(z), -1) = \lim_{z \rightarrow -1} (z+1) \cdot \frac{5z^2 - 4z + 3}{(z+1)(z+2)(z+3)} = \lim_{z \rightarrow -1} \frac{5z^2 - 4z + 3}{(z+2)(z+3)} = 6. \]
3Step 3: Find \( \operatorname{Res}(f(z), -2) \)
For the pole \( z = -2 \), calculate: \[ \operatorname{Res}(f(z), -2) = \lim_{z \rightarrow -2} (z+2) \cdot \frac{5z^2 - 4z + 3}{(z+1)(z+2)(z+3)} = \lim_{z \rightarrow -2} \frac{5z^2 - 4z + 3}{(z+1)(z+3)} = -31. \]
4Step 4: Find \( \operatorname{Res}(f(z), -3) \)
Similar to previous steps, identify that \( z = -3 \) is a pole and compute: \[ \operatorname{Res}(f(z), -3) = \lim_{z \rightarrow -3} (z+3) \cdot \frac{5z^2 - 4z + 3}{(z+1)(z+2)(z+3)} = \lim_{z \rightarrow -3} \frac{5z^2 - 4z + 3}{(z+1)(z+2)} = 30. \]
Key Concepts
Residue TheoremPoleRational FunctionLimit Calculation
Residue Theorem
The Residue Theorem is a powerful tool in complex analysis, particularly useful for evaluating certain kinds of integrals. It connects the mathematics of residues at poles of meromorphic functions with contour integrals. In simple terms, the theorem states that the integral of a function around a closed contour is related to the sum of the residues of the function inside the contour. Specifically, if the function is analytic except for isolated singularities within the contour, the integral is equal to \( 2\pi i \times \) the sum of the residues at those singularities.
The formula for computing residues, which is essential for applying the theorem, involves taking the limit of the product of \((z-a)\) and the function \(f(z)\), where \(a\) is a point of singularity. The residue at a simple pole \(a\) is given by:
This method is straightforward when the singularities are simple poles, which leads us to our next concept, poles.
The formula for computing residues, which is essential for applying the theorem, involves taking the limit of the product of \((z-a)\) and the function \(f(z)\), where \(a\) is a point of singularity. The residue at a simple pole \(a\) is given by:
- \( \operatorname{Res}(f(z), a) = \lim_{z \to a} (z-a) f(z) \)
This method is straightforward when the singularities are simple poles, which leads us to our next concept, poles.
Pole
In complex analysis, a pole of a function is a type of singularity that resembles a "spike" in the function's value. It's a point \(a\) where a function \(f(z)\) goes to infinity. Poles are characterized by the behavior of the function as it approaches some finite value, or infinity, and they can be of varying orders.
A simple pole, which we focus on here, occurs when \((z-a)f(z)\) is analytic and non-zero at \(z = a\). This means the function sharply rises or falls to infinity as it nears the pole, but it does so linearly with respect to \((z-a)\).
Identifying these poles is essential when calculating residues since they are often the points of evaluation for these calculations. In the provided example, the poles are \(-1\), \(-2\), and \(-3\) within the rational function given.
A simple pole, which we focus on here, occurs when \((z-a)f(z)\) is analytic and non-zero at \(z = a\). This means the function sharply rises or falls to infinity as it nears the pole, but it does so linearly with respect to \((z-a)\).
Identifying these poles is essential when calculating residues since they are often the points of evaluation for these calculations. In the provided example, the poles are \(-1\), \(-2\), and \(-3\) within the rational function given.
Rational Function
A rational function is one of the simplest forms of a complex function, represented as the quotient of two polynomials. It takes the form \(f(z) = \frac{P(z)}{Q(z)}\), where both \(P(z)\) and \(Q(z)\) are polynomials. The behavior of the rational function, especially near its poles, is critical in complex analysis.
The denominator, \(Q(z)\), must not be zero for the function to be defined. If \(Q(z) = 0\) at point \(z = a\), then \(a\) is a pole of the function. The degree of these zeros helps determine the order of the pole.
In problems like the one outlined, it's crucial to express the rational function accurately to identify where the poles are. This clarity simplifies the application of the residue theorem, as it allows for straightforward limit calculations at these poles.
The denominator, \(Q(z)\), must not be zero for the function to be defined. If \(Q(z) = 0\) at point \(z = a\), then \(a\) is a pole of the function. The degree of these zeros helps determine the order of the pole.
In problems like the one outlined, it's crucial to express the rational function accurately to identify where the poles are. This clarity simplifies the application of the residue theorem, as it allows for straightforward limit calculations at these poles.
Limit Calculation
Limit calculation is a fundamental process in complex analysis, particularly when working with residues. When calculating limits in complex analysis, we assess the behavior of a function as it approaches a particular point.
When a function has a simple pole at \(z = a\), the residue at this pole can be calculated using the limit:
This removes the indeterminate form and isolates the residue, as seen in the examples for the poles \(-1, -2,\) and \(-3\) from the original exercise. Mastering limit calculations is key to effectively using the residue theorem in complex analysis.
When a function has a simple pole at \(z = a\), the residue at this pole can be calculated using the limit:
- \( \operatorname{Res}(f(z), a) = \lim_{z \to a} (z-a) f(z) \)
- Identify the pole \(a\) in the rational function.
- Multiply the function \(f(z)\) by \((z-a)\).
- Take the limit as \(z\) approaches \(a\).
This removes the indeterminate form and isolates the residue, as seen in the examples for the poles \(-1, -2,\) and \(-3\) from the original exercise. Mastering limit calculations is key to effectively using the residue theorem in complex analysis.
Other exercises in this chapter
Problem 10
From \(\quad f(z)=z-\sin z=\frac{z^{3}}{3 !}-\frac{z^{5}}{5 !}+\cdots=z^{3}\left(\frac{1}{3 !}-\frac{z^{2}}{5 !}+\cdots\right)\) we see that \(z=0\) is a zero o
View solution Problem 10
Replacing \(z\) in \(\sin z=\sum_{k=0}^{\infty}(-1)^{k} \frac{z^{2 k+1}}{(2 k+1) !}\) by \(3 z\) gives \(\sin 3 z=\sum_{k=0}^{\infty}(-1)^{k} \frac{(3 z)^{2 k+1
View solution Problem 11
$$\operatorname{Re}\left(z_{n}\right)=\frac{8 n^{2}+n}{4 n^{2}+1} \rightarrow 2 \text { as } n \rightarrow \infty, \text { and } \operatorname{Im}\left(z_{n}\ri
View solution Problem 12
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
View solution