# Complex Analysis - Essential Singularities and Poles

• Dan7620
In summary, the solution for finding two analytic functions f and g with a common essential singularity at z=0, but a product function f(z)g(z) with a pole of order 5 at z=0 is to use the functions f(z) = e^(1/z) and g(z) = (1/z^5)(e^(-1/z)). Their respective Laurent series centered around 0 will have an infinite number of terms with negative powers of z, but their product will have a finite number of terms with 5 negative powers of z, resulting in a pole of order 5 at z=0.
Dan7620

## Homework Statement

Find two analytic functions f and g with common essential singularity at z=0, but the product function f(z)g(z) has a pole of order 5 at z=0.

## Homework Equations

Not an equation per say, but I'm thinking of the desired functions in terms of their respective Laurent series centered around 0 (which exist since the functions are analytic.)

## The Attempt at a Solution

1. I know that any function with an essential singularity at z=0 will have an infinite number of negative powers of z when expressed as a Laurent Series.

2.I also know that if the product function has a pole of order 5, then its Laurent series will have a finite number terms in negative powers of z (...5 terms max?).

With this knowledge, I wrote out the general Laurent series of f(z) and g(z), found their product (convolution of the series) and am now trying to find two functions f(z) and g(z) such that the resultant series has 5 terms with negative powers of z.

Needless to say, I'm stuck and I would greatly appreciate any input here. Thanks for your time.

It might be easier if you think of a function f(z) which has an essential singularity at z=0 and then concoct a function g(z) so that the essential part of the singularity cancels in the product.

OK I believe I've solved the problem... both these functions f and g should have essential singularities at z=0, and their product has a pole of order 5 at zero:

$$f\left( z \right)=e^{\frac{1}{z}}$$
$$g\left( z \right)=\frac{1}{e^{\frac{1}{z}}z^{5}}$$

Would anyone mind confirming this? Thanks

Can I confirm it? Or are you feeling insecure?

I'm a little doubtful because this solution seems trivial, but for the sake of completeness, here is my work:

$$f\left( z \right)=\sum_{n\; =\; 0\; }^{\infty }{\frac{1}{n!z^{n}}}$$

$$g\left( z \right)=\frac{1}{z^{5}}\left( \sum_{n\; =\; 0}^{\infty }{\frac{\left( -\frac{1}{z} \right)^{n}}{n!}} \right)=\sum_{n\; =\; 0}^{\infty }{\frac{\left( -1 \right)^{n}}{n!z^{n+5}}}$$

and clearly the product function
$$f\left( z \right)g\left( z \right)=\frac{1}{z^{5}}$$

Looking at the Laurent series expressions for f and g it shows that they both have essential singularities at z=0 (b/c of the infinite number of terms with negative powers of z), while the product has a pole of order 5.

Could you confirm this?

Thanks

Dan7620 said:
I'm a little doubtful because this solution seems trivial, but for the sake of completeness, here is my work:

$$f\left( z \right)=\sum_{n\; =\; 0\; }^{\infty }{\frac{1}{n!z^{n}}}$$

$$g\left( z \right)=\frac{1}{z^{5}}\left( \sum_{n\; =\; 0}^{\infty }{\frac{\left( -\frac{1}{z} \right)^{n}}{n!}} \right)=\sum_{n\; =\; 0}^{\infty }{\frac{\left( -1 \right)^{n}}{n!z^{n+5}}}$$

and clearly the product function
$$f\left( z \right)g\left( z \right)=\frac{1}{z^{5}}$$

Looking at the Laurent series expressions for f and g it shows that they both have essential singularities at z=0 (b/c of the infinite number of terms with negative powers of z), while the product has a pole of order 5.

Could you confirm this?

Thanks

I think that's just fine.

## 1. What is a singularity in complex analysis?

A singularity in complex analysis refers to a point in the complex plane where a function is undefined or behaves in an unusual way. This can include essential singularities, poles, and removable singularities.

## 2. What is an essential singularity?

An essential singularity is a type of singularity where a function has infinite oscillations or jumps near the singularity point. In other words, the function cannot be extended to that point in a smooth manner. This type of singularity is the most severe and can be identified by analyzing the behavior of the function as it approaches the singularity point.

## 3. What is a pole in complex analysis?

A pole is a type of singularity where a function approaches infinity as it gets closer to the singularity point. This can be seen as a vertical asymptote on a graph of the function. Poles can be further classified as simple poles or higher order poles, depending on the degree to which the function approaches infinity at the singularity point.

## 4. How do essential singularities and poles affect the behavior of a function?

Essential singularities and poles can significantly impact the behavior of a function in the complex plane. They can cause the function to have infinite values or oscillations, as well as affect the convergence or divergence of the function in specific regions. Understanding the location and type of singularity can provide insight into the behavior of the function and its properties.

## 5. Can essential singularities and poles be avoided in complex analysis?

In general, essential singularities and poles cannot be avoided since they occur when a function is undefined or behaves unexpectedly. However, in some cases, it may be possible to manipulate equations or use certain techniques to avoid or remove singularities, such as using the theory of residues to integrate around a singularity point.

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