What Is the Laurent Series for e^(1/z)?

[Physgeek64]

Homework Statement

Cassify the singularities of e^\frac{1}{z} and find the Laurent series

Homework Equations

e^\frac{1}{x}=\sum \frac{(\frac{1}{x})^n}{n!}

The Attempt at a Solution

there's a singularity at z=0, but I need to find the order of the pole

So using the general expression for the expansion of an exponential:
e^\frac{1}{z}=\sum \frac{(\frac{1}{z})^n}{n!} but this leads to a 1 as the first term, which is obviously not consistent.

I also tried considering re-defining a new variable for \frac{1}{z}, but I'm not really sure how to proceed from here

Many thanks :)

 

[stevendaryl]

Homework Statement

Cassify the singularities of e^\frac{1}{z} and find the Laurent series

Homework Equations

e^\frac{1}{x}=\sum \frac{(\frac{1}{x})^n}{n!}

The Attempt at a Solution

there's a singularity at z=0, but I need to find the order of the pole

So using the general expression for the expansion of an exponential:
e^\frac{1}{z}=\sum \frac{(\frac{1}{z})^n}{n!} but this leads to a 1 as the first term, which is obviously not consistent.

If the order of the pole for f(z) at z=0 is n, then that means that f(z) \propto \frac{1}{z^n} near z=0, which in turn means that z^n f(n) remains finite as z \rightarrow 0. So in the case of e^{\frac{1}{z}}, do you think there is some number n that would work?

 

[Physgeek64]

If the order of the pole for f(z) at z=0 is n, then that means that f(z) \propto \frac{1}{z^n} near z=0, which in turn means that z^n f(n) remains finite as z \rightarrow 0. So in the case of e^{\frac{1}{z}}, do you think there is some number n that would work?

Im sorry I don't see how you concluded that f(z) \propto \frac{1}{z^n} near z=0, since would the expansion for f(z) =1-\frac{1}{z}+\frac{1}{2!} \frac{1}{z^2}+...

Thank you :)

 

[BvU]

Steven didn't conclude it. He wrote If ... then ... !

 

[stevendaryl]

Im sorry I don't see how you concluded that f(z) \propto \frac{1}{z^n} near z=0, since would the expansion for f(z) =1-\frac{1}{z}+\frac{1}{2!} \frac{1}{z^2}+...

Let me quote from Wikipedia (https://en.wikipedia.org/wiki/Singularity_(mathematics)#Complex_analysis)

• The point a is a removable singularity of f if there exists a holomorphic function g defined on all of U such that f(z) = g(z) for all z in U \ {a}. The function g is a continuous replacement for the function f.
• The point a is a pole or non-essential singularity of f if there exists a holomorphic function g defined on U with g(a) nonzero, and a natural number n such that f(z) = g(z) / (z − a)n for all z in U \ {a}. The number n here is called the order of the pole. The derivative at a non-essential singularity itself has a non-essential singularity, with n increased by 1 (except if n is 0 so that the singularities are removable).
• The point a is an essential singularity of f if it is neither a removable singularity nor a pole. The point a is an essential singularity if and only if the Laurent series has infinitely many powers of negative degree.

 

[Physgeek64]

Im sorry I don't see how you concluded that f(z) \propto \frac{1}{z^n} near z=0, since would the expansion for f(z) =1-\frac{1}{z}+\frac{1}{2!} \frac{1}{z^2}+...

Thank you :)

Let me quote from Wikipedia (https://en.wikipedia.org/wiki/Singularity_(mathematics)#Complex_analysis)

• The point a is a removable singularity of f if there exists a holomorphic function g defined on all of U such that f(z) = g(z) for all z in U \ {a}. The function g is a continuous replacement for the function f.
• The point a is a pole or non-essential singularity of f if there exists a holomorphic function g defined on U with g(a) nonzero, and a natural number n such that f(z) = g(z) / (z − a)n for all z in U \ {a}. The number n here is called the order of the pole. The derivative at a non-essential singularity itself has a non-essential singularity, with n increased by 1 (except if n is 0 so that the singularities are removable).
• The point a is an essential singularity of f if it is neither a removable singularity nor a pole. The point a is an essential singularity if and only if the Laurent series has infinitely many powers of negative degree.

Ah okay, so in this case we have an essential singularity?

Would I be okay in thinking that the Laurent series is f(z) =1-\frac{1}{z}+\frac{1}{2!} \frac{1}{z^2}+... ? Or have I oversimplified the problem

Thank you :)

 

[Physgeek64]

Steven didn't conclude it. He wrote If ... then ... !

My mistake, I misinterpreted his response. Thank you for clarifying :)