Math Problem Statement

also evaluate its residue points

Solution

To determine the poles and their orders for the given function f(z)f(z), and to evaluate the residues at those points, we proceed as follows:

The function is given by: f(z)=z1(z+1)2(z2)f(z) = \frac{z-1}{(z+1)^2 (z-2)}

1. Finding the Poles and Their Orders

Poles are the values of zz for which the denominator of f(z)f(z) is zero, provided that the numerator does not also become zero at those points.

The denominator of f(z)f(z) is (z+1)2(z2)(z+1)^2 (z-2). Setting this equal to zero gives us: (z+1)2(z2)=0(z+1)^2 (z-2) = 0

So, the poles are at: z=1andz=2z = -1 \quad \text{and} \quad z = 2

Order of the Poles:

  • At z=1z = -1: The factor (z+1)2(z+1)^2 indicates a pole of order 2.
  • At z=2z = 2: The factor (z2)(z-2) indicates a pole of order 1.

2. Evaluating the Residues

The residue of a function f(z)f(z) at a pole z0z_0 is given by:

  • For a simple pole (order 1): Residue=limzz0(zz0)f(z)\text{Residue} = \lim_{z \to z_0} (z - z_0) f(z)
  • For a pole of order mm: Residue=1(m1)!limzz0dm1dzm1[(zz0)mf(z)]\text{Residue} = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z - z_0)^m f(z) \right]

Residue at z=1z = -1 (pole of order 2):

Residue=11!limz1ddz[(z+1)2f(z)]\text{Residue} = \frac{1}{1!} \lim_{z \to -1} \frac{d}{dz} \left[ (z + 1)^2 f(z) \right] (z+1)2f(z)=(z1)z2(z + 1)^2 f(z) = \frac{(z-1)}{z-2} ddz[(z1)z2]=(z2)1(z1)1(z2)2=(z2)(z1)(z2)2=1(z2)2\frac{d}{dz} \left[ \frac{(z-1)}{z-2} \right] = \frac{(z-2) \cdot 1 - (z-1) \cdot 1}{(z-2)^2} = \frac{(z-2) - (z-1)}{(z-2)^2} = \frac{-1}{(z-2)^2} Evaluating this at z=1z = -1: Residue=11!(1(z2)2)z=1=1(12)2=19=19\text{Residue} = \frac{1}{1!} \left. \left( \frac{-1}{(z-2)^2} \right) \right|_{z = -1} = \frac{-1}{(-1-2)^2} = \frac{-1}{9} = -\frac{1}{9}

Residue at z=2z = 2 (pole of order 1):

Residue=limz2(z2)f(z)\text{Residue} = \lim_{z \to 2} (z - 2) f(z) (z2)f(z)=(z2)(z1)(z+1)2(z2)=(z1)(z+1)2(z - 2) f(z) = \frac{(z-2)(z-1)}{(z+1)^2 (z-2)} = \frac{(z-1)}{(z+1)^2} Evaluating this at z=2z = 2: Residue=(z1(z+1)2)z=2=21(2+1)2=19\text{Residue} = \left. \left( \frac{z-1}{(z+1)^2} \right) \right|_{z = 2} = \frac{2-1}{(2+1)^2} = \frac{1}{9}

Summary

  • The function f(z)f(z) has poles at z=1z = -1 and z=2z = 2.
    • The pole at z=1z = -1 is of order 2.
    • The pole at z=2z = 2 is of order 1.
  • The residues are:
    • Residue at z=1z = -1 is 19-\frac{1}{9}.
    • Residue at z=2z = 2 is 19\frac{1}{9}.

Would you like any further details or have any other questions?

Related Questions

  1. How do you find the poles of a rational function?
  2. What is the definition of a residue in complex analysis?
  3. How do you determine the order of a pole?
  4. What is the significance of a residue at a pole?
  5. Can a function have an essential singularity? How is it different from a pole?
  6. How do you compute residues for higher-order poles?
  7. What are Laurent series and how do they relate to residues?
  8. How can residues be used to evaluate complex integrals?

Tip

Understanding how to compute residues and the behavior of poles is essential for evaluating complex integrals using the residue theorem.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Complex Analysis
Poles
Residues

Formulas

Residue formula for simple and higher-order poles

Theorems

Residue theorem

Suitable Grade Level

Advanced Undergraduate