Laurent series of e^(1/z)

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

Theres 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
Staff Emeritus

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

Theres 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?

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

stevendaryl
stevendaryl
Staff Emeritus
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) / (za)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.

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) / (za)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 :)

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

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