Math Problem Statement

((log(z))/(z^(3)+1)) find residues

Solution

To find the residues of the function

f(z)=log(z)z3+1f(z) = \frac{\log(z)}{z^3 + 1}

we need to follow these steps:

1. Find the poles of the function:

The residues are located at the poles of the function. The poles occur when the denominator is zero, i.e., when:

z3+1=0z^3 + 1 = 0

Solving for zz, we get:

z3=1z^3 = -1

The solutions to this equation are the cube roots of -1. These roots are:

z=eiπ/3,z=eiπ,z=ei5π/3z = e^{i\pi/3}, \quad z = e^{i\pi}, \quad z = e^{i5\pi/3}

More explicitly, the poles are:

z1=1,z2=ei2π/3,z3=ei4π/3z_1 = -1, \quad z_2 = e^{i2\pi/3}, \quad z_3 = e^{i4\pi/3}

2. Classify the poles:

Since z3+1=0z^3 + 1 = 0 is a cubic equation, each of these poles is a simple pole. We can see this by noting that the derivative of z3+1z^3 + 1 is non-zero at these points.

3. Find the residues:

To compute the residues at each pole zkz_k, we use the formula for the residue of a simple pole:

Res(f,zk)=limzzk(zzk)f(z)\text{Res}(f, z_k) = \lim_{z \to z_k} (z - z_k) f(z)

That is, we need to evaluate:

Res(f,zk)=limzzk(zzk)log(z)z3+1\text{Res}(f, z_k) = \lim_{z \to z_k} \frac{(z - z_k)\log(z)}{z^3 + 1}

The function z3+1z^3 + 1 can be factored as:

z3+1=(zzk)(zz2)(zz3)z^3 + 1 = (z - z_k)(z - z_2)(z - z_3)

Thus, the residue at each zkz_k simplifies to:

Res(f,zk)=log(zk)3zk2\text{Res}(f, z_k) = \frac{\log(z_k)}{3z_k^2}

Now, compute the residues at each pole:

  • At z1=1z_1 = -1: Res(f,1)=log(1)3(1)2=iπ3\text{Res}(f, -1) = \frac{\log(-1)}{3(-1)^2} = \frac{i\pi}{3}

  • At z2=ei2π/3z_2 = e^{i2\pi/3}: Res(f,ei2π/3)=log(ei2π/3)3(ei2π/3)2=i2π/33ei4π/3\text{Res}(f, e^{i2\pi/3}) = \frac{\log(e^{i2\pi/3})}{3(e^{i2\pi/3})^2} = \frac{i2\pi/3}{3e^{i4\pi/3}} Simplifying, we get: Res(f,ei2π/3)=i2π/9ei4π/3\text{Res}(f, e^{i2\pi/3}) = \frac{i2\pi/9}{e^{i4\pi/3}}

  • At z3=ei4π/3z_3 = e^{i4\pi/3}: Res(f,ei4π/3)=log(ei4π/3)3(ei4π/3)2=i4π/33ei8π/3\text{Res}(f, e^{i4\pi/3}) = \frac{\log(e^{i4\pi/3})}{3(e^{i4\pi/3})^2} = \frac{i4\pi/3}{3e^{i8\pi/3}} Simplifying further: Res(f,ei4π/3)=i4π/9ei8π/3\text{Res}(f, e^{i4\pi/3}) = \frac{i4\pi/9}{e^{i8\pi/3}}

Conclusion:

The residues of log(z)z3+1\frac{\log(z)}{z^3 + 1} at the poles z1=1z_1 = -1, z2=ei2π/3z_2 = e^{i2\pi/3}, and z3=ei4π/3z_3 = e^{i4\pi/3} are:

  • Res(f,1)=iπ3\text{Res}(f, -1) = \frac{i\pi}{3}
  • Res(f,ei2π/3)=i2π/9ei4π/3\text{Res}(f, e^{i2\pi/3}) = \frac{i2\pi/9}{e^{i4\pi/3}}
  • Res(f,ei4π/3)=i4π/9ei8π/3\text{Res}(f, e^{i4\pi/3}) = \frac{i4\pi/9}{e^{i8\pi/3}}

Would you like further details or clarification on any steps?

Related Questions:

  1. How do you compute residues for higher-order poles?
  2. What is the importance of the residue theorem in complex analysis?
  3. How does the logarithmic branch cut affect the evaluation of residues?
  4. Can we apply contour integration to find the sum of residues?
  5. What are the applications of computing residues in physics?

Tip:

When computing residues at roots of unity, consider using polar coordinates for simplifications involving logarithms and powers of complex numbers.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Complex Analysis
Residues
Poles
Logarithmic Functions

Formulas

Residue formula for a simple pole: Res(f, z_k) = lim_{z -> z_k} (z - z_k) f(z)
z^3 + 1 = (z - z_1)(z - z_2)(z - z_3)
Residue at z_k = log(z_k) / (3z_k^2)

Theorems

Residue Theorem
Pole classification

Suitable Grade Level

Undergraduate - Complex Analysis