What is the difference between regular and simple pole?

Example $f(z)$ - regular

For example, the function $f(z)=\frac{z}{z^2+1}$ has a regular point at infinity. I can show this by finding the residue at:$$\frac{1}{z^2}\cdot f\left(\frac{1}{z}\right)= \frac{1}{z(z+1)}\approx \frac{1}{z}\left[ 1 - z +z^2 \right].$$ The residue for this is the opposite of the coefficient of the term $z^{-1}$. In this example it is $\frac{1}{z}$, giving us a residue of -1 at $z_0=0$.

Example $g(z)$ - regular

The function $g(z)=\sin(\frac{1}{z})$ has a regular point at infinity. Expanding this: $$\frac{1}{z^2}\cdot g\left(\frac{1}{z}\right) = \frac{\sin(z)}{z^2}\approx \frac{1}{z^2}\left[ z - \frac{z^3}{3!} \right]= \frac{1}{z} - \frac{z}{3!},$$ with the residue at zero is -1 again.

Example $h(z)$ - simple pole

The function $h(z)=\frac{4z^3+2z +3}{z^2}$ has a simple pole at infinity. I do not need the Laurent series for this because it is already in a simple form or already a series: $$\frac{1}{z^2}\cdot h\left(\frac{1}{z}\right)= \frac{4}{z^3}+\frac{2}{z} + 3 $$


Why does $f(z), g(z)$ have a regular pole and $h(z)$ have a simple pole?


There are two misunderstandings here.

First, there is no such thing as a "regular pole". I believe you are referring to "regular point", which is any point where $f$ is well-defined and holomorphic. In other words, a "regular point" is any point that is NOT a pole or an essential singularity!

Second, to determine whether the point at infinity is a pole or a regular point of a function $f(z)$, you should be looking at the Laurent series for the function $f(1/w)$ around $w = 0$. Not the Laurent series for $w^{-2}f(1/w)$.

[We want to think of the point at infinity as the point $z = \infty$ - except this doesn't exist in the space parametrised by the $z$ coordinate. So we introduce a new coordinate $w = 1/z$. The point at infinity is then represented by $w = 0$.]

For example, if $f(z) = \sin (1/z)$, then $$f(1/w) = \sin w = w - \frac 1 6 w^3 + \dots ,$$ which is regular at $w = 0$, since the Laurent series has no negative powers of $w$.

However, if $f(z) = \frac{4z^3+2z + 3}{z^2}$, then $$f(1/w) = 4/w + 2+ 3w^2,$$ which has a single pole at $w = 0$.

Having said that, one situation where you do need to consider $w^{-2} f(1/w)$ is when you are computing a line integral, $\int_C f(z) dz$. This is because the integration measure $dz$ becomes $dz = -dw/w^2$ in the $w$ coordinates, giving you the extra factor of $w^{-2}$.

  • $\begingroup$ P.S. For the example $f(z) = z / (z^2 + 1)$, your expression for $f(1/w)$ isn't right. You should get $f(1/w) = w/(1+w^2) = w - w^3 + w^5 - w^7 + \dots$. And yes, this is regular at $w = 0$, because its series expansion contains no negative powers of $w$. $\endgroup$ – Kenny Wong May 15 '17 at 0:56

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