Complex Analysis - Essential Singularities and Poles

[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.

 

[Dick]

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.

 

[Dan7620]

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:

<br /> f\left( z \right)=e^{\frac{1}{z}}<br />
<br /> g\left( z \right)=\frac{1}{e^{\frac{1}{z}}z^{5}}<br />


Would anyone mind confirming this? Thanks

 

[Dick]

Can I confirm it? Or are you feeling insecure?

 

[Dan7620]

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

<br /> f\left( z \right)=\sum_{n\; =\; 0\; }^{\infty }{\frac{1}{n!z^{n}}}<br />

<br /> 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}}}<br />

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

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

 

[Dick]

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

<br /> f\left( z \right)=\sum_{n\; =\; 0\; }^{\infty }{\frac{1}{n!z^{n}}}<br />

<br /> 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}}}<br />

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

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.