Finding a Laurent series / residue problem

Ocifer
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Homework Statement


f(z) = \frac{1}{ \exp{ \frac {z^2 - \pi/2}{ \sqrt{3} } } + i }

Find the residue of f(z) at z_0 = \frac{ \sqrt(\pi) }{2 } ( \sqrt(3) - i )


Homework Equations



The Attempt at a Solution



I was able to verify that the given z_0 is a singularity, and furthermore an essential singularity. However, I am stumped at how to figure out the residue (value of a_{-1} ) from the given information.

I've tried rearranging letting f(z) be a general Laurent series, and then rearranged the equation. I've tried making multiplicative arguments, like below:

\sum_{n \in Z} d_n (z - z_0)^n \cdot ( \exp{ \frac {z^2 - \pi/2}{ \sqrt{3} } } + i ) = 1

In the past I've been able to figure out the residue from multiplicative arguments and matching powers, but I don't see how I can do that here. Since the given z_0 is an essential singularity, there will necessarily be an infinite number of negative-degree terms, which must cancel with the infinite number of positive-degree terms from the exponential portion, and it gets so messy that I can't resolve anything.

Furthermore, I am troubled by the fact that I would need an infinite number of negative powers to cancel with an infinite number of positive powers, and so I am not even sure if the usual approach of matching powers will terminate or give an answer.

Can anyone provide a hint, or point out an error I may have made?

I should mention that I've found series for the exponential expression and trivially for the 1, both about z0. But I'm not finding a nice pattern for the coefficients of the exponential so that I can equate coefficients. The first few terms I have but they're not pretty
 
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Ocifer said:

Homework Statement


f(z) = \frac{1}{ \exp{ \frac {z^2 - \pi/2}{ \sqrt{3} } } + i }

Find the residue of f(z) at z_0 = \frac{ \sqrt(\pi) }{2 } ( \sqrt(3) - i )


Homework Equations



The Attempt at a Solution



I was able to verify that the given z_0 is a singularity, and furthermore an essential singularity. However, I am stumped at how to figure out the residue (value of a_{-1} ) from the given information.

I've tried rearranging letting f(z) be a general Laurent series, and then rearranged the equation. I've tried making multiplicative arguments, like below:

\sum_{n \in Z} d_n (z - z_0)^n \cdot ( \exp{ \frac {z^2 - \pi/2}{ \sqrt{3} } } + i ) = 1

In the past I've been able to figure out the residue from multiplicative arguments and matching powers, but I don't see how I can do that here. Since the given z_0 is an essential singularity, there will necessarily be an infinite number of negative-degree terms, which must cancel with the infinite number of positive-degree terms from the exponential portion, and it gets so messy that I can't resolve anything.

Furthermore, I am troubled by the fact that I would need an infinite number of negative powers to cancel with an infinite number of positive powers, and so I am not even sure if the usual approach of matching powers will terminate or give an answer.

Can anyone provide a hint, or point out an error I may have made?

I should mention that I've found series for the exponential expression and trivially for the 1, both about z0. But I'm not finding a nice pattern for the coefficients of the exponential so that I can equate coefficients. The first few terms I have but they're not pretty

It doesn't look like an essential singularity to me. It looks like a simple pole. Why do you think it's essential? Did you try just computing the limit z->z0 (z-z0)*f(z)?
 
Thank you for pointing that out, I must have been careless earlier. Using L'Hopital's rule on the indeterminate "0/0" form, I also now get that it is a simple pole. After that I used a result about Laurent series and residue about a pole of order m.

Thank you.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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