Question

Find and classify the isolated singularities of each of the following functions and compute the residues of the singularities.

f(z) = (z³)(e¹/ᶻ)

Answer 1

The function
`\(f(z) = z^3 \cdot e^(1/z)\)` has an essential singularity at z = 0. The residue at this singularity is 1, determined from the 1/z term in the Laurent series expansion.

To find and classify the isolated singularities of
`\(f(z) = z^3 \cdot e^(1/z)\)`, we need to analyze the behavior of the function around points where it is not defined.

The function f(z) has an isolated singularity at z = 0 due to the term
`\(e^(1/z)\)`. Let's examine the nature of this singularity:

1. (z = 0):

- This is an essential singularity because the function has an infinite number of terms in its Laurent series expansion. Specifically, the term
`\(e^(1/z)\)` has an infinite series expansion around z = 0.

Now, let's compute the residue at z = 0. The residue is given by the coefficient of the 1/z term in the Laurent series expansion. However, since the function has an essential singularity at z = 0, finding the exact Laurent series expansion can be challenging.

In general, the residue at an essential singularity is not as straightforward to compute as for poles. It often involves more advanced techniques or specific methods depending on the function. In this case, finding the residue at z = 0 requires expanding the function in a Laurent series, and the 1/z term in that series would be the residue.

`\[ f(z) = z^3 \cdot e^(1/z) = z^3 \cdot \left(1 + (1)/(z) + (1)/(2z^2) + (1)/(6z^3) + \ldots\right) \]`

The residue at z = 0 is the coefficient of the 1/z term, which is 1.

In summary:

- z = 0 is an essential singularity.

- The residue at z = 0 is 1.