Understanding the Principal Part of the Function ( frac{1 - cos{z}}{z^2} ) and Its Residue

The function ( f(z) frac{1 - cos{z}}{z^2} ) is an interesting example when discussing singularities and their classification in complex analysis. In this detailed explanation, we will explore the nature of the singularity at ( z 0 ), compute its Laurent series expansion, and discuss the principal part and residue of this function.

Singularities and Their Classification

To determine the type of singularity at ( z 0 ), we first need to identify any potential singularities. Since the numerator and denominator are entire functions, the only possible singularities occur when the denominator equals zero. In this case:

( z 0 ) is the only singularity of the given function.

Computing the Laurent Series Expansion

The main idea to simplify this process is to use the standard Maclaurin expansion for the cosine function: (cos{z} sum_{n0}^{infty} frac{(-1)^n z^{2n}}{(2n)!} ). Let's compute the Laurent series expansion for ( f(z) ) at ( z 0 ).

( f(z) frac{1 - cos{z}}{z^2} frac{1 - left(1 - sum_{n0}^{infty} frac{(-1)^n z^{2n}}{(2n)!}right)}{z^2} frac{1}{z^2} left(1 - 1 sum_{n0}^{infty} frac{(-1)^n z^{2n}}{(2n)!}right) frac{1}{z^2} cdot sum_{n1}^{infty} frac{(-1)^n z^{2n}}{(2n)!} sum_{n1}^{infty} frac{(-1)^{n - 1} z^{2n - 2}}{(2n)!} sum_{n0}^{infty} frac{(-1)^n z^{2n}}{2(n 1)!} ) via re-indexing.

The Principal Part and Removable Singularity

The Laurent series expansion of ( f(z) ) at ( z 0 ) is:

( f(z) sum_{n0}^{infty} frac{(-1)^n z^{2n}}{2(n 1)!} )

We can observe that there are no nonzero terms with negative powers of ( z ). This means that there is no principal part in the Laurent series, as all the terms have non-negative powers of ( z ).

Since ( f(z) ) does not have any principal part, the singularity at ( z 0 ) is classified as a removable singularity. Consequently, the residue of this function at ( z 0 ) is 0.

Conclusion

In conclusion, the function ( f(z) frac{1 - cos{z}}{z^2} ) has a removable singularity at ( z 0 ). The computation of the Laurent series showed that there are no terms with negative powers of ( z ), and hence the residue is 0.

Key Points

The singularity of ( f(z) frac{1 - cos{z}}{z^2} ) at ( z 0 ) is removable. The principal part of the function does not exist. The residue of the function at ( z 0 ) is 0.

Understanding these concepts is fundamental in complex analysis and is crucial for proper classification and computation of singularities in various functions.