Understanding the Residue of ( frac{1}{z^3 - z^5} ) at ( z infty )

In this article, we will explore how to calculate the residue of the function ( frac{1}{z^3 - z^5} ) at infinity. This is a critical concept in complex analysis, particularly in the context of residue theorem and its applications.

1. Residue Theorem and the Definition of Residue at Infinity

Residue theory is a fundamental concept in complex analysis, allowing us to evaluate contour integrals using the residues of the function within the contour. A key aspect of residue theory is understanding how residues behave at infinity. The definition of the residue of a function at infinity is crucial for our argument. According to the definition:

(text{Res}f(z)_{infty} -text{Res}frac{1}{z^2} fleft(frac{1}{z}right)_{0})

This definition transforms the problem at infinity into a problem at zero, which is often more straightforward to solve.

2. Applying the Definition to ( frac{1}{z^3 - z^5} )

Consider the function:

(f(z) frac{1}{z^3 - z^5})

To find the residue at infinity, we first need to rewrite the function in the form of (frac{1}{z^2} fleft(frac{1}{z}right)). Substituting (z) with (frac{1}{z}), we get:

(fleft(frac{1}{z}right) frac{1}{left(frac{1}{z}right)^3 - left(frac{1}{z}right)^5} frac{z^5}{1 - z^2})

Thus, the function (frac{1}{z^2} fleft(frac{1}{z}right)) becomes:

(frac{1}{z^2} cdot frac{z^5}{1 - z^2} frac{z^3}{1 - z^2})

3. Analysis of the Function at ( z 0 )

Now, let's analyze the function (g(z) frac{z^3}{1 - z^2}) at (z 0).

[Inline math] [ g(z) frac{z^3}{1 - z^2} ]

First, observe that at (z 0), the function (g(z)) has a triple zero, which indicates that the function has a zero of order 3 at (z 0). This means that the function is well-behaved and does not have a pole or any other non-removable singularity at (z 0).

Since the function has a triple zero at (z 0), the worst that can happen is a removable singularity. This implies that the residue at (z 0) for the function (frac{1}{z^2} fleft(frac{1}{z}right)) is zero.

Hence, the residue of ( f(z) frac{1}{z^3 - z^5} ) at (z infty) is:

(text{Res}_{zinfty} f(z) -text{Res}_{z0} frac{1}{z^2} fleft(frac{1}{z}right) -0 0)

4. Conclusion

In conclusion, we have shown that the residue of the function ( frac{1}{z^3 - z^5} ) at infinity is zero. This is achieved by transforming the problem using the residue theorem and analyzing the behavior of the function at the origin. This approach highlights the power and utility of the residue theorem in complex analysis.