Residue of ${ z }^{ 3 }\cosh { \frac { 1 }{ z } } $ What is the residue of ${ z }^{ 3 }\cosh { \frac { 1 }{ z }  } $ at $z=0$?
I have that:
${ z }^{ 3 }\cosh { \frac { 1 }{ z }  } =1+\frac { z }{ 2! } +\frac { { z }^{ -1 } }{ 4! } +\frac { { z }^{ -3 } }{ 6! } +...$
The ${ z }^{ -1 }$ term of the Laurent Series is $\frac { 1 }{ 4! } $. I think that this implies that the residue of the given function at $z=0$ is $\frac { 1 }{ 4! } $.
Wolframalpha indicates that ${ z }^{ 3 }\cosh { \frac { 1 }{ z }  } $ does not have a pole at $z=0$. Am I incorrect?
 A: You are correct about the residue. In fact, though, it doesn't have a pole at $z=0$. Rather, it has an essential singularity. This is because the principal part of the Laurent series has infinitely-many non-zero terms. If there were a pole, then there would be only finitely-many such terms.
A: In complex analysis, there is more than one type of singularity.  Suppose $f(x)$ is holomorphic in a punctured neighborhood of $z_0$. Then these things may happen:


*

*$f$ actually can be extended to $z_0$ analytically.  Consider $f(z) = \frac{z}{z}$ at $z=0$.  This is called a removable singularity.

*$f$ has a singularity, but $f$ can be written as a quotient of two holomorphic functions, with the denominator zero at $z_0$.  Consider $\frac{z}{z^2 + 1}$ at $i$, or $z^{-3}$ at $0$.  This is called a pole.  If you treat $f$ as a function from a neighborhood $z_0$ to the Riemann Sphere, it is an $n$ to $1$ map from the neighborhood to a neighborhood of $\infty$.  Functions of this sort are called meromorphic.

*Neither case happens.  There is no $k$ such that $(z-z_0)^k f(z)$ can be extended to $z_0$ analytically.  This is called an essential singularity.  $e^{\frac{1}{z}}$ at $z = 0$ is a great example.  Such singularities have special properties.  The Casorati-Weierstraß theorem states that $f$ takes any punctured neighborhood of $z_0$ to a dense subset of $\mathbf{C}$.  The much harder Big Picard Theorem states that $f$ takes any neighborhood of $z_0$ either to $\mathbf{C}$ or to $\mathbf{C}$ with one point removed.

