If the only tool you have is a hammer, you tend to see every problem as a nail, Abraham Maslow.

Removable singularities

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.

Examples

The Removable Singularity, $f_1(z) = \frac{e^z-1}{z}$ At z = 0, this looks like $\frac{0}{0}$, which is undefined. If we look at the Taylor series expansion of $e^z: e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \dots = \frac{e^z - 1}{z} = \frac{(1 + z + \frac{z^2}{2!} + \dots) - 1}{z} = \frac{z + \frac{z^2}{2!} + \dots}{z} = 1 + \frac{z}{2!} + \dots$. As $z \to 0$, the limit clearly approaches 1. Therefore, the “singularity” is just a missing point. We can “plug the hole” by defining f(0) = 1, $f_1(x) = \begin{cases} \frac{e^z-1}{z}, &z \ne 0 \\\\ 1, &z = 0 \end{cases}$, making the function perfect (analytic).
The Pole, $f_2(z) = \frac{1}{z} \text{ or } \frac{1}{z^n}$, As $z \to 0$, the modulus $|f_2(z)|$ shoots to infinity; no matter which direction you approach the singularity 0 from, the value gets infinitely large. However, we can “fix” this by multiplying by a factor of $z$ (or $z^n$). $\lim_{z \to 0} z \cdot \frac{1}{z} = 1, (\lim_{z \to 0} z \cdot \frac{1}{z} = 1\lim_{z \to 0} z^n \cdot \frac{1}{z^n} = 1)$. This is a pole. It cannot be removed, but it behaves predictably (it goes to $\infty$).
The Essential Singularity, $f_3(z) = e^{1/z}$. The function behavior is quite chaotic: (i) If z approaches 0 from the positive real axis ($z \to 0^+$), $1/z = 1/x \to \infty$, so $e^{1/z} = e^{1/x} \to \infty$, so the function blows up; (ii) If z approaches 0 from the negative real axis ($z \to 0^-$), $1/z = 1/x \to -\infty$, so $e^{1/z} = e^{1/x} \to 0$, so here it collapses to zero; (iii) If z approaches along the imaginary axis (z = iy, $y \to 0$), $\frac{1}{z} = \frac{1}{iy} = \frac{-i}{y}, e^{1/z} = e^{-i/y} = cos(\frac{-1}{y})+sin(\frac{-1}{y}) = cos(\frac{1}{y})-sin(\frac{-1}{y})$ which has modulus 1 and oscillates wildly around the unit circle as $y\rightarrow 0$. The limit does not exist (it is neither finite nor infinity). This is an essential singularity, aka the “worst” kind.

Intuitive definition. A removable singularity of a function f at a point a is an isolated singularity where f(z) can be redefined at the point a to make the function analytic there. The function “wants” to be analytic. The singularity is just a definition gap or a missing point, not a fundamental geometric feature of the function.

Definition. A function f is said to have a removable singularity at a point a if f is analytic in a punctured disk B'(a; r) and there exists an analytic function g on the full disk B(a; r) that agrees with f on the punctured disk, $g(z) = f(z), \forall z \in B'(a, r)$ In other words, there exists a “valid, nice, or cool version” g of our original function that fills in the gap seamlessly.

Lemma. Vanishing Integral under “Mild” Singularity. Let f be analytic on the punctured disk $\mathbb{B'}(a; r) = \{z \in \mathbb{C} : 0 < |z - a| < r\}$. If f satisfies the condition $\lim_{z \to a}(z - a)f(z) = 0$, then $\int_{\gamma} f(\xi)d\xi = 0$ for any simple closed curve $\gamma$ in $\mathbb{B'}(a; r)$. This condition $\lim_{z \to a}(z - a)f(z) = 0$ says as z approaches a, the function f(z) grows strictly slower than $\frac{1}{z-a}$, the prototype of a simple pole (a pole of order 1). f is allowed to blow up at a, but not as fast as a simple pole.

Proof.

We need to calculate $\int_{\gamma} f(z)dz$. We distinguish two cases based on the position of the singularity a relative to the curve $\gamma$.

Case 1: The singularity $a$ is outside $\gamma$. Let $I(\gamma)$ denote the interior region bounded by $\gamma$. If $a \notin I(\gamma)$, then the entire region $I(\gamma)$ lies within the punctured disk $\mathbb{B}'(a; r)$ where f is analytic. The function f is analytic on $\gamma$ and everywhere inside it. By the Cauchy-Goursat Theorem, the integral vanishes : $\int_{\gamma} f(z)dz = 0$

Recall. Cauchy-Goursat Theorem. If a function f is analytic on a simply connected domain D and $\gamma$ is any simple closed contour lying entirely within D, then $\oint_{\gamma} f(z) dz = 0$.

Case 2: The singularity a is inside $\gamma$. If $a \in I(\gamma)$, we cannot use the standard Cauchy Theorem directly.

We are given the condition: $\lim_{z \to a}(z - a)f(z) = 0$. By the formal definition of a limit, for any arbitrary $\varepsilon > 0$, there exists a small radius $\delta > 0$ such that: $|z - a||f(z)| \lt \frac{\varepsilon}{2\pi}, \forall z \in \mathbb{B}(a; \delta)$.

Consider a small circle $\mathbb{C}_{\delta₁}$ centered at a of radius $\delta_1, 0 \lt \delta_1 \lt \delta$. We choose $\delta_1$ small enough so that this circle lies entirely inside $\gamma$. The function f is analytic in the region between $\gamma$ and $C_{\delta_1}$. By the Principle of Deformation of Paths, the integral over the original curve equals the integral over this small circle: $\int_{\gamma} f(z)dz = \int_{C_{\delta_1}} f(z)dz$

Recall. Principle of Deformation of Paths. If f is analytic in a region containing two simple closed loops $\gamma_1$ and $\gamma_2$, and the region between them contains no singularities, then: $\oint_{\gamma_1} f(z) dz = \oint_{\gamma_2} f(z) dz$.

Now we estimate the integral over the small circle $C_{\delta_1}$ using the ML-Inequality (Estimation Lemma). On the circle $C_{\delta_1}$, we have $|z - a| = \delta_1 \lt \delta$.

$$ \begin{aligned} \left| \int_{\gamma} f(z)dz \right| &=\left| \int_{C_{\delta_1}} f(z)dz \right| \\[2pt] &\le \int_{C_{\delta_1}} |f(z)| |dz| \\[2pt] &\lt[|z-a| \lt \delta, \text{we can substitute the previous bound}] \int_{C_{\delta_1}} \frac{\varepsilon}{2\pi |z - a|} |dz| \\[2pt] &=[|z - a| = \delta_1] \frac{\varepsilon}{2\pi \delta_1} \int_{C_{\delta_1}} |dz| \\[2pt] &=[\text{The integral } \int_{C_{\delta_1}} |dz| \text{ is the circumference (length) of the circle}] \frac{\varepsilon}{2\pi \delta_1} \cdot (2\pi \delta_1) = \varepsilon \end{aligned} $$

Since the magnitude of the integral $|\int_{\gamma} f(\xi)d\xi|$ is arbitrarily small (or smaller than any positive number), it must be exactly zero $\int_{\gamma} f(\xi)d\xi = 0$.

Remark. Generalization to Multiple Singularities. We can generalize or extend this result to a function with finitely many singularities $a_1, a_2, \cdots, a_n$. If f is analytic on a disk except at these points $\mathbb{B}(a; r) \setminus \{ a_1, a_2, \cdots, a_n \}$ and the limit condition $\lim_{z \to a_i}(z-a_i)f(z) = 0$ holds for each $a_i$ (each singularity is removable in the sense of vanishing residude), then $\int_{\gamma} f(\xi)d\xi = 0$ for any simple closed contour $\gamma$ in the punctured domain $\mathbb{B}(a; r) \setminus \{ a_1, a_2, \cdots, a_n \}$.

Idea. It relies on the Deformation of Paths. Imagine a rubber band (our contour $\gamma$) wrapping around a region containing several “holes” (singularities).

We can pinch the rubber band together (or deform the large contour $\gamma$) to isolate each hole into a set of tiny disjoints circles $C_1, C_2, \dots, C_n$, where each circle $C_i$ surrounds exactly one singularity $a_i$. Then, $\oint_{\gamma} f(\xi)d\xi = \sum_{i=1}^n \oint_{C_i} f(\xi)d\xi$. The intermediate paths from the rubber band or large contour $\gamma$ to each $C_i, i = 1, \cdots, n$ cancel because each deformation path is traversed once forward and once backward.

We could apply the previous Lemma (Single Singularity case) to each integral in the sum. Since $\lim_{z \to a_i}(z-a_i)f(z) = 0$ for every i, we know that the total integral vanishes $\oint_{C_i} f(\xi)d\xi = 0$.

Lemma. Cauchy Integral Formula with a “Mild” Singularity. Let f be analytic in the punctured disk $\mathbb{B'}(a; r)$ and satisfy $\lim_{z \to a}(z - a)f(z) = 0$. Let $\gamma$ be a circle centered at a with radius $r_1, 0 \lt r_1 \lt r$. Then, for any z inside $\gamma$ (where $z \ne a$): $f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi -z} d\xi, \forall z \in \mathbb{B'}(a; r_1)$

This lemma is important and powerful. It states that even if a function is undefined at a point a, as long as the singularity is “mild” (limit is 0), the Cauchy Integral Formula still works or holds as if the singularity wasn’t there. In other words, f(z) can be represented by a Cauchy Integral.

Proof.

Fix a point $z_0 \in \mathbb{B'}(a; r_1)$. We define a specific difference quotient, $F(z) = \frac{f(z) - f(z_0)}{z - z_0}$.

Analyze the Singularities of F. At z = $z_0$, the denominator is zero (this is a singularity introduced by our formula). At z = a, the function f itself has a singularity by assumption. So, F(z) has two singularities inside $\gamma$: one at $z_0$ and one at a.

To be able use the Generalization Remark (proving that $\oint F dz = 0$), we must show that F satisfies the limit or vanishing condition at both singularities.

$\lim_{z \to z_0} F(z)(z - z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}\cdot (z - z_0) = \lim_{z \to z_0} f(z) - f(z_0) = 0$ since f is continuous at $z_0, f(z) \to f(z_0)$, hence condition satisfied at $z_0$.

$\lim{z \to a} F(z)(z - a) = \lim_{z \to a}\frac{f(z) - f(z_0)}{z - z_0}\cdot (z - a) = \lim_{z \to a} \frac{f(z)(z-a)}{z-z_0} -\lim_{z \to a} \frac{f(z_0)(z - a)}{z - z_0} = 0 -0 = 0$.

Term 1. The denominator $z - z_0 \to a - z_0 \ne 0$. The numerator $f(z)(z -a) \to 0$ by the hypothesis of the Lemma. Thus, Term 1 $\to \frac{0}{a-z_0} = 0$.
Term 2. The denominator remains the same, $z - z_0 \to a - z_0 \ne 0$. The numerator $f(z_0)(z-a) \to f(z_0)\cdot 0 = 0$. Thus, Term 2 $\to 0$.

Now we can safely apply the previously proved Generalization Remark. Since F(z) is an analytic function in $\gamma$ except at a and $z_0$, and also satisfies the limit conditions at both points, the integral of F around $\gamma$ must vanish: $\oint_\gamma F(z) dz = 0$.

Next, we substitute the definition of F back into the integral, $\int_\gamma \frac{f(z) - f(z_0)}{z - z_0}dz = 0 \implies[\text{Split the integral using linearity:}] \int_\gamma \frac{f(z) }{z - z_0}dz = f(z_0)\int_\gamma \frac{1}{z - z_0}dz = f(z_0) 2\pi i$ (this last equality holds because $z_0$ is inside $\gamma$).

Then, $f(z_0) = \frac{1}{2\pi i}\int_\gamma \frac{f(z) }{z - z_0}dz$. Since $z_0 \in \mathbb{B'}(a; r_1)$ was chosen arbitrarily, then the lemma holds.

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.

Proof.

Necessity. Assume there exists a function g that is analytic on the whole disk $\mathbb{B}(a; r)$ and $g(z) = f(z)$ for all $z \neq a$, then the limit condition must hold.

Since g is analytic at a, it is continuous at a. This means $\lim_{z \to a} g(z) = g(a)$, a finite number.

We need to compute the limit in question, $\lim_{z \to a}(z - a)f(z) =[\text{Since } g(z) = f(z), \forall z \in \mathbb{B'}(a; r)] \lim_{z \to a}(z - a)g(z) =[\text{Using the product rule for limits}] \left( \lim_{z \to a} (z - a) \right) \cdot \left( \lim_{z \to a} g(z) \right) = 0 \cdot g(a) = 0$.

Suppose there were two analytic extensions, $g_1$ and $g_2$. They both equal $f(z)$ on the punctured disk $\mathbb{B}'(a; r)$. Since the punctured disk is a set with a limit point, the Identity Theorem tells us that if two analytic functions agree there, they must agree everywhere in the domain. Thus, $g_1 = g_2$.

The set $\mathbb{B}'(a;r)$ is open and dense in $\mathbb{B}(a;r)$. In particular, every point of the disk —including the center a— is a limit point of $\mathbb{B}'(a;r)$.

Sufficient. If the limit is zero $\lim_{z \to a}(z - a)f(z) = 0$, then we can construct an analytic extension g.

Recall the Cauchy Integral Formula with a “Mild” Singularity. Under this exact condition, for any circle $C_{r_1}$ inside the domain, the Cauchy Integral Formula holds for $f(z)$ inside the circle (excluding $a$): $f(z) = \frac{1}{2\pi i}\int_{C_{r_1}} \frac{f(\xi)}{\xi -z}d\xi$ where $C_{r_1}$ is a circle centered at a with radius $r_1, 0 \lt r_1 \lt r$ and $z \in \mathbb{B'}(a; r_1)$.

Define our candidate extension, $g(z) = \frac{1}{2\pi i} \oint_{C_{r_1}} \frac{f(\xi)}{\xi - z} d\xi \quad \text{for ALL } z \in \mathbb{B}(a; r_1)$

Suppose $z \neq a$. By the Cauchy Integral Formula with a “Mild” Singularity, this integral equals $f(z)$. So $g(z) = f(z)$ and g is analytic because it equals f.

Suppose z = a. This integral gives us the specific value needed to plug the “hole” (fake singularity). We only need to prove it is differentiable at the trouble point z = a. We use the definition of the derivative.

Let h be a small complex number. We want to show the limit of the difference quotient $\frac{g(a+h) - g(a)}{h}$ does exist.

$$ \begin{aligned} \frac{g(a+h) - g(a)}{h} &= \frac{1}{h} \left[ \frac{1}{2\pi i} \oint_{C_{r_1}} \frac{f(\xi)}{\xi - (a+h)} d\xi - \frac{1}{2\pi i} \oint_{C_{r_1}} \frac{f(\xi)}{\xi - a} d\xi \right] \\[2pt] &\text{Combine the integral} \\[2pt] &= \frac{1}{2\pi i h} \oint_{C_{r_1}} \left( \frac{f(\xi)}{\xi - a - h} - \frac{f(\xi)}{\xi - a} \right) d\xi \\[2pt] &\text{Algebraic Simplification (common denominator)}: \\[2pt] &\frac{1}{\xi - a - h} - \frac{1}{\xi - a} = \frac{(\xi - a) - (\xi - a - h)}{(\xi - a - h)(\xi - a)} = \frac{h}{(\xi - a - h)(\xi - a)} \\[2pt] &= \frac{1}{2\pi i \not{h}} \oint_{C_{r_1}} f(\xi) \frac{\not{h}}{(\xi - a - h)(\xi - a)} d\xi \end{aligned} $$

Now we take the limit $\lim_{h \to 0} \frac{g(a+h)-g(a)}{h} \frac{1}{2\pi i \not{h}} \oint_{C_{r_1}} f(\xi) \frac{\not{h}}{(\xi - a - h)(\xi - a)} d\xi$. As $h \to 0$, the term $(\xi - a - h)$ becomes $(\xi - a)$. Since the integration path $C_{r_1}$ is a fixed distance away from a, we can safely pass the limit inside the integral (Uniform Convergence).

Note: Swapping the limit and the integral symbols is forbidden if the function blows up (becomes infinite) somewhere during the limit process.

However, the term we are integrating is $\frac{f(\xi)}{(\xi - a - h)(\xi - a)}$. This fraction would explode if the denominator became zero. That happens if $\xi = a$ ($\xi$ live on the circle $C_{r_1}$ and their distance from the center a is exactly $r_1$, so $\xi \ne a$) or if $\xi = a+h$. Could $\xi$ ever equal $a+h$?

We are taking the limit as $h \to 0$ (a + h is the perturbation, near the center a). We can restrict our view to very small h. Let’s say we only care about $h$ such that $|h| < \frac{r_1}{2}$. The distance from any point $\xi$ on the rim of the circle to the center a is $r_1$. By the Triangle Inequality, $|\xi - (a+h)| \ge |\xi - a| - |h| = r_1 - |h| \ge r_1 -\frac{r_1}{2} = \frac{r_1}{2}$, hence the whole fraction is uniformly bounded and cannot spike to infinity anywhere on the integration path.

$\lim_{h \to 0} \frac{g(a+h) - g(a)}{h} = \frac{1}{2\pi i} \oint_{C_{r_1}} \frac{f(\xi)}{(\xi - a)^2} d\xi$

This integral is over a closed, bounded (compact) curve (the circle $C_{r_1}$). $\xi$ represents points on the circle, a is the center of the circle, so $|\xi - a|$ is exactly equal to the radius $r_1$ everywhere on the path and is obviously never zero; hence the denominator is never zero on the path. Since f is analytic on the circle (away from the center), $f(\xi)$ is continuous.

We are integrating a continuous function over a finite curve. We know that a continuous function on a compact set is bounded. The integral of a bounded function over a finite length is always a finite number.

Since this integral exists and is finite, the derivative $g'(a)$ exists. Therefore, $g$ is analytic at $a$. Since $g$ agrees with $f$ everywhere else, $g$ is the required analytic extension.

Removable Singularities

Removable singularities

Examples