Poles Classification of Isolated Singularities

There are two ways to do great mathematics. The first is to be smarter than everybody else. The second way is to be stupider than everybody else — but persistent, Raoul Bott.

Recall

Definition. We say that a point $a \in \mathbb{C}$ is a regular point of a function f if f is analytic at a. This is a “good” or “nice” point. The function is smooth, differentiable, and behaves well here.

Definition. We say that a point $a \in \mathbb{C}$ is a singularity of a function f if f is not analytic at a, but every neighborhood of a contains points where f is analytic. This is a “bad” point. The function crashes, explodes/blows up, or is undefined here, but it is right on the edge of “good” territory.

The function behaves beautifully on a punctured neighborhood, but at the center point, something fundamentally fails.

Definition. Suppose f is analytic on a punctured disk $B'(a; r) = \{ z : 0 < |z-a| < r \}$. Then, we say f has an isolated singularity at a.

Complex analysis classifies isolated singularities into three specific types based on how the function behaves as it gets close to the singularity.

Riemann’s Removable Singularity. Suppose f(z) is analytic on the punctured disk $\mathbb{B'}(a; r)$. There exists a unique analytic function g(z) on the full disk $\mathbb{B}(a; r)$ so that $g(z) = f(z), \forall z \in \mathbb{B'}(a; r)$ if and only if $\lim_{z \to a}(z - a)f(z) = 0.$ In other words, (1) we have a hole or singularity in the domain at z = a; (2) The limit condition $\lim_{z \to a}(z - a)f(z) = 0$ means that even if f(z) goes to infinity at $a$, it does so “relatively slowly” (at least, slower than a pole of order 1, like $1/z$); (iii) If this condition holds, the singularity is fake, not real, and the function can be redefined at the point a in order to make it analytic (smooth everywhere) in the whole disk.

Often we will write this g(a) as f(a) and call g to be the function f with the singularity at a removed.

We start with a function f(z) that shoots to infinity as z gets close to a. Let f be analytic on $\mathbb{B}'(a; r)$ with $\lim_{z \to a}f(z) = \infty$.

Using the formal definition of a limit approaching infinity. Given any arbitrary “height” M > 0 (no matter how big), $\exist \delta \gt 0, \text{ s.t., } |f(z)| \gt M, \forall z \in \mathbb{B}'(a; \delta)$ (if we get close enough to a, the function’s value is higher than the height).

If we pick a specific height, say M = 1, we know there is a small neighborhood around a ($\delta_1$) where $|f(z)| > 1 > 0$. This guarantees that we can divide by f(z) without dividing by zero in $\mathbb{B}'(a; \delta_1)$.

Since $f(z) \ne 0$ near a, we define the reciprocal function $g(z) = \frac{1}{f(z)}$. g is well defined and analytic in $\mathbb{B}'(a; \delta_1)$. Furthermore, the singularity of g at a is removable since $\lim_{z \to a}(z-a)g(z) = 0$.

Because f goes to infinity, g goes to zero, $\lim_{z \to a} g(z) = \lim_{z \to a} \frac{1}{f(z)} = 0.$

Because $\lim_{z \to a} g(z)$ exists and is finite (it is indeed 0), the singularity at a is removable (fake). We “fill in the hole” by defining g(a) = 0 (we just rename g to its analytic extension for simplicity). Now g(z) is analytic everywhere in the disk.

Now we have an analytic function g(z) that is zero at a. How “strong” is this zero? In complex analysis, zeroes are integers. The function might behave like $(z-a)$ (order 1), or $(z-a)^2$ (order 2), etc. We assume g has a zero of order k. Since g is analytic in a neighborhood of a, it has a Taylor series: $g(z)=\sum _{n=0}^{\infty }c_n(z-a)^n.$ This allows us to factor out the “zero part”: $g(z) = (z-a)^k \cdot g_1(z)$ where $g_1(z)$ is the “leftover” part of the function that is not zero at a and $g_1$ is analytic on $\mathbb{B}(a; \delta_2)$

We want to know about f, not g. So we substitute g = 1/f back into our equation: $\frac{1}{f(z)} = (z-a)^k g_1(z), \forall z \in \mathbb{B}'(a; \delta_2) \implies f(z)(z-a)^k = \frac{1}{g_1(z)}$

$\lim_{z \to a} f(z)(z-a)^k = \lim_{z \to a} \frac{1}{g_1(z)} =[g_1 \text{ is analytic on } \mathbb{B}(a; \delta_2)] \frac{1}{g_1(a)} \ne 0$

This gives us the Structure of a Pole: $f(z) \approx \frac{\text{Constant}}{(z-a)^k}$

What happens when we multiply f(z) by different powers of $(z-a)$. In other words, we are looking for the “Goldilocks” power that cancels the infinity exactly.

1. The Perfect Match ($\alpha = k$). If $\lim_{z \to a} f(z)(z-a)^k = \dots \frac{1}{g_1(a)} \ne 0$.
2. Power is too high ($\alpha > k$, then we have “extra” zeroes on top). $\lim_{z \to a} f(z)(z-a)^{\alpha} = (\lim_{z \to a} f(z)(z-a)^{\alpha-k}) \cdot (\lim_{z \to a} f(z)(z-a)^{k}) = 0 \cdot \text{Finite} = 0$
3. Power is too low ($\alpha < k$). $\lim_{z \to a} f(z)(z-a)^{\alpha} = \lim_{z \to a} \frac{f(z)(z-a)^k}{(z-a)^{k -\alpha}} = \infty$ (we don’t have enough power to cancel the denominator completely. We are left with terms in the denominator, so the limit still shoots to infinity.)

In this case, we say that f has a pole of order k at a. Conclusion: $\lim_{z \to a}(z-a)^kf(z) \ne 0 \iff f$ has a pole of order k at a.

Let’s provide a final “sanity check” or intuitive verification for the conclusion already reached. Let’s show that if the limit condition holds, the function f(z) must behave like a pole of order m.

Assume the limit condition is true for some integer m: $\lim_{z \to a} (z-a)^m f(z) = D$ where D is a constant and $D \neq 0$.

If a function approaches a limit D, then close to the limit point, the function equals D plus a tiny “error” term. Let $h(z) = (z-a)^m f(z)$. Since by assumption $h(z) \to D$, then using the definition of a limit we can rewrite is as $h(z) = D + w(z)$ where D is the non-zero constant limit and w(z) is the error term. As $z \to a$, the “remainder” term w(z) vanishes ($w \to 0$).

$(z-a)^m f(z) = D + w(z) \implies f(z) =[\text{Solving for f(z)}] \frac{D + w(z)}{(z-a)^m}$

$lim_{z \to a}f(z) = lim_{z \to a}\frac{D + w(z)}{(z-a)^m}$

• Numerator: As $z \to a$, $w(z) \to 0$, so the numerator approaches $D + 0 = D$. Since $D \neq 0$, the numerator is a stable, non-zero number.
• Denominator: As $z \to a$, $(z-a)^m \to 0$.

Therefore, we have a non-zero number divided by a number approaching zero, $lim_{z \to a}f(z) = lim_{z \to a}\frac{D + w(z)}{(z-a)^m} = \frac{\text{Non-Zero Constant D}}{\text{Tiny Number} \to 0} = \infty$. This confirms that f(z) indeed goes to infinity at $a$, and the rate at which it goes to infinity is driven exactly by the power m in the denominator. Thus, f has a pole of order m at a.