I'm trying to understand how my book determines the poles of a function and their corresponding order, by just looking at the terms in the denominator and not using Laurent series. Hope somebody can clarify this.

My book teached me that:

Singularities are isolated points where the complex derivative of $f(z)$ doesn't exist.

The most negative power of the term $(z-z_0)$ in the Laurent expansion of $f(z)$ about the singularity $z_0$ is some finite power $(z-z_0)^{-n}$. Then $z_0$ is a pole of order $n$.

But after explaining what a pole is, the book started finding poles and the corresponding order of the pole, just by factoring the denominator of a given function. So I just want to understand why it is that simple.

Consider a function $f(z)$ that has been rewritten as $$f(z)=\frac{z^4+1}{z^2(z-\frac{1}{2})(z-2)}.$$ Then the book determines immediately that the function has poles at $z=0$ of order $2$, $z=1/2$ and $z=2$ of order $1$.

The first issue is determining the singularities. It can easily be seen that the function diverges at these points and thus the derivative can't exist at these points. But is it always this easy? I believe you can construct functions where points of divergence is not the only types of singularities? But anyway, finding singularities is not really what I'm concerned about and it's pretty easy in this case.

After you have found the singularities. How can you tell that the power of the singularity in the denominator corresponds directly to the most negative term in the Laurent Series expanded about that point? In other words, how can you easily see that $1/(z-2)$ will be the highest negative power in the Laurent expansion around $z=2$, and thus $z_0=2$ is a pole of first order?

Is this having something to do with partial fraction decomposition of the denominator?:

$$f(z)=(z^4+1)(\frac{A}{z} +\frac{B}{z^2} +\frac{C}{z-\frac{1}{2}}+\frac{D}{z-2})$$

Because I can't quite see how that would help.

  • $\begingroup$ I am troubled by your definition of "singularity". Has $f(z):=\frac{1}{z}$ whose derivative is $-\frac{1}{z^2}$ got a singularity at $z=0$ according to your definition? $\endgroup$ Commented Feb 21, 2021 at 12:08
  • $\begingroup$ @ancientmathematician Yes. Is it not a singularity? $\endgroup$
    – sjm23
    Commented Feb 21, 2021 at 12:15
  • $\begingroup$ Yes, of course, but I don't really see why you bring the derivative into the story. $\endgroup$ Commented Feb 21, 2021 at 12:20
  • $\begingroup$ @Ancientmathematician Because my book says that a singularity is isolated points where the function is not analytic, which means the derivative doesn't exist at the given point? $\endgroup$
    – sjm23
    Commented Feb 21, 2021 at 12:22
  • $\begingroup$ You should probably just ignore my comments then. My puzzle is whether $f$ itself has to be defined at the singularity. $\endgroup$ Commented Feb 21, 2021 at 12:30

1 Answer 1


Note that, if$$g(z)=\frac{z^4+1}{z^2(z-1/2)},$$then$$f(z)=\frac{g(z)}{z-2}.$$But $g$ is analytic near $2$ and $g(2)\ne0$. So, near $2$, you can write $g(z)$ as$$g(z)=a_0+a_1(z-2)+a_2(z-2)^2+a_3(z-2)^3+\cdots,$$ with $a_0\ne0$, and therefore$$f(z)=\frac{a_0}{z-2}+a_1z+a_2(z-2)+a_3(z-2)^2+\cdots$$So, indeed, $f$ has a pole of order $1$ at $2$. A similar argument works for the poles of $f$ at $0$ and at $\frac12$.

  • $\begingroup$ Thank you! That makes a lot of sense. $\endgroup$
    – sjm23
    Commented Feb 21, 2021 at 12:14

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