Zeros, singularities, and residues

A function that blows up is not automatically a mystery — complex analysis gives a complete, three-way classification of every point where an analytic function misbehaves, and attaches to each such point a single extractable number, the residue, that will later do the work of an entire contour integral. The classification comes for free once you have the Laurent series about the point: look only at its negative-power terms, the principal part, and every isolated singularity sorts itself into removable, pole, or essential. Zeros get a matching, dual classification by order. This article builds both classifications from the Laurent series you already met, derives the shortcut formulas that let you find a residue without ever writing out a full series, and works a Laurent-based example alongside the formula-based one to show the two methods agree. It sets up the Residue Theorem, the topic immediately following.

The core insight

Every isolated singularity of an analytic function is classified — removable, a pole of order $m$ , or essential — by how many negative-power terms sit in its Laurent series' principal part, and the residue, the coefficient $c_{-1}$ , is the one number from that series that survives contour integration.

The core idea

Zeros and their order. A point $z_0$ is a zero of order $m$ of an analytic function $f$ if $f(z_0) = f'(z_0) = \cdots = f^{(m-1)}(z_0) = 0$ but $f^{(m)}(z_0) \ne 0$ . Equivalently, near $z_0$ the function factors as

f(z) = (z-z_0)^m g(z),

where: $g$ is analytic at $z_0$ and $g(z_0) \ne 0$ — the order $m$ is exactly the power of $(z-z_0)$ you can pull out before what remains stops vanishing. A zero of order $1$ is called a simple zero. This factored form is the mirror image of the pole formula below, and the two are used together constantly: a zero of $q(z)$ in a denominator is precisely what creates a pole of $f = p/q$ .

Isolated singularities and the Laurent test. $z_0$ is an isolated singularity of $f$ if $f$ is analytic on a punctured disk $0 < |z-z_0| < r$ but not analytic (or not even defined) at $z_0$ itself. Expand $f$ there as a Laurent series,

f(z) = \sum_{n=-\infty}^{\infty} c_n (z-z_0)^n,

where: the terms with $n<0$ form the principal part. Reading only that principal part sorts every isolated singularity into exactly one of three types. A removable singularity has principal part identically zero — every $c_n=0$ for $n<0$ — so $\lim_{z\to z_0}f(z)$ exists and defining $f(z_0)$ to equal it makes $f$ analytic there; $f(z)=\sin z/z$ at $z=0$ is the standard example. A pole of order $m$ has a principal part that stops after finitely many terms, with $c_{-m}\ne 0$ and every $c_n=0$ for $n<-m$ ; equivalently $f(z) = h(z)/(z-z_0)^m$ with $h$ analytic and $h(z_0)\ne 0$ . A pole of order $1$ is a simple pole. An essential singularity has infinitely many nonzero negative-power terms, as in $f(z)=e^{1/z}$ at $z=0$ , whose series $\sum_{n\ge 0} \frac{1}{n!\,z^n}$ never terminates going leftward. The figure below lays these three cases out as bars along the Laurent index $n$ , making the "how far left do the nonzero bars extend" test visual.

Figure

Bar chart of Laurent series coefficients against index n from minus three to three showing a removable singularity with no bars left of zero, a pole of order two with exactly two bars left of zero, and an essential singularity with bars continuing indefinitely to the left

Residue. The residue of $f$ at an isolated singularity $z_0$ , written $\operatorname{Res}_{z=z_0}f(z)$ , is defined as the Laurent coefficient $c_{-1}$ . It is the only coefficient that matters for contour integration because $\oint_C (z-z_0)^n\,dz = 0$ for every integer $n\ne -1$ around a small loop about $z_0$ , but equals $2\pi i$ when $n=-1$ — so integrating the whole Laurent series term by term kills every term except the $c_{-1}$ one. A removable singularity always has residue $0$ , since its principal part vanishes entirely.

Worked example

Problem. Find the residue of $f(z) = \dfrac{e^z-1}{z^3}$ at $z=0$ , first by expanding a Laurent series directly and then by the order- $m$ pole formula, and confirm the two answers agree.

Step 1. Expand the numerator with the known Maclaurin series $e^z = 1+z+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}+\dfrac{z^4}{4!}+\cdots$ , so $e^z-1 = z+\dfrac{z^2}{2}+\dfrac{z^3}{6}+\dfrac{z^4}{24}+\cdots$ .

Step 2. Divide by $z^3$ term by term: $f(z) = \dfrac{1}{z^2}+\dfrac{1}{2z}+\dfrac{1}{6}+\dfrac{z}{24}+\cdots$ . The principal part has exactly two nonzero terms, $1/z^2$ and $1/(2z)$ , so $z=0$ is a pole of order $2$ , and reading off the coefficient of $1/z$ gives $\operatorname{Res}_{z=0}f(z) = \dfrac{1}{2}$ directly from the series.

Step 3. Cross-check with the order- $m$ formula (derived below) for $m=2$ : $\operatorname{Res}_{z=0}f(z) = \lim_{z\to 0}\dfrac{d}{dz}\!\left[z^2 f(z)\right] = \lim_{z\to 0}\dfrac{d}{dz}\!\left[\dfrac{e^z-1}{z}\right]$ . Writing $\dfrac{e^z-1}{z} = 1+\dfrac{z}{2}+\dfrac{z^2}{6}+\cdots$ from Step 1, its derivative is $\dfrac{1}{2}+\dfrac{z}{3}+\cdots$ , which at $z=0$ gives $\dfrac{1}{2}$ . Both routes agree. The residue of $f(z) = (e^z-1)/z^3$ at $z=0$ is $\dfrac{1}{2}$ , and $z=0$ is a pole of order $2$ .

Competitive stretch

The same function also has an effective order lower than the denominator's bare factor of $3$ suggests, precisely because the numerator itself vanishes at $z=0$ (a simple zero, since $e^z-1 \approx z$ there): the pole order is $3-1=2$ , not $3$ . Spotting this cancellation before applying any residue formula is exactly the habit tested in the traps below.

Common traps

Assuming every singularity is a pole. Check the Laurent expansion, or at least the vanishing/non-vanishing of a limit, before reaching for a pole-residue formula — essential and removable singularities need entirely different treatment, and the pole formulas give nonsense if misapplied to them.
Misjudging pole order from the denominator's factored form alone. If the numerator also vanishes at that point, as in the worked example above, the effective order is lower than the denominator's exponent suggests; always confirm the numerator does not cancel part of the singularity first.
Forgetting the $\frac{1}{(m-1)!}$ factor or differentiating the wrong number of times. For a pole of order $m$ you differentiate $m-1$ times and divide by $(m-1)!$ , not $m$ times — an easy off-by-one slip.
Treating the residue of a removable singularity as nonzero. By definition it is always $0$ , since there is no principal part to contribute a $1/z$ term.

Derivation AThe residue formula for a pole of order m

The goal is to derive, purely from the definition of the pole (a Laurent series whose principal part terminates after $m$ terms) and ordinary differentiation, the shortcut formula $\operatorname{Res}_{z=z_0}f(z) = \frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}\big[(z-z_0)^m f(z)\big]$ that lets you extract a residue without ever writing out a full Laurent series by hand.

Step 1. Start from the definition of a pole of order $m$ : near $z_0$ , $f(z) = \dfrac{c_{-m}}{(z-z_0)^m} + \dfrac{c_{-m+1}}{(z-z_0)^{m-1}} + \cdots + \dfrac{c_{-1}}{z-z_0} + c_0 + c_1(z-z_0) + \cdots$ , a Laurent series whose principal part has exactly $m$ terms. The target coefficient is $c_{-1}$ .

Step 2. Multiply through by $(z-z_0)^m$ to clear every negative power at once: $(z-z_0)^m f(z) = c_{-m} + c_{-m+1}(z-z_0) + \cdots + c_{-1}(z-z_0)^{m-1} + c_0(z-z_0)^m + \cdots$ . This is now an ordinary Taylor series in $(z-z_0)$ , analytic at $z_0$ itself, because every negative power has been absorbed.

Step 3. Recall how Taylor coefficients are extracted: if $\phi(z) = \sum_{k\ge 0} a_k (z-z_0)^k$ , then $a_k = \dfrac{\phi^{(k)}(z_0)}{k!}$ . Apply this to $\phi(z) := (z-z_0)^m f(z)$ from Step 2, whose coefficient of $(z-z_0)^{m-1}$ is exactly $c_{-1}$ , the residue. So $c_{-1} = \dfrac{\phi^{(m-1)}(z_0)}{(m-1)!}$ .

Step 4. Write $\phi^{(m-1)}(z_0)$ as a limit, since $\phi$ is analytic there and derivatives are continuous: $\phi^{(m-1)}(z_0) = \lim_{z\to z_0}\dfrac{d^{m-1}}{dz^{m-1}}\big[(z-z_0)^m f(z)\big]$ . Combining with Step 3 gives the stated formula, $\operatorname{Res}_{z=z_0}f(z) = \dfrac{1}{(m-1)!}\lim_{z\to z_0}\dfrac{d^{m-1}}{dz^{m-1}}\big[(z-z_0)^m f(z)\big]$ . Setting $m=1$ collapses this to the simple-pole case, $\operatorname{Res}_{z=z_0}f(z) = \lim_{z\to z_0}(z-z_0)f(z)$ , with no derivative needed at all — consistent with $0! = 1$ and a zeroth derivative being the function itself.

Ask the tutor: If distinguishing pole order from the numerator or denominator's structure trips you up, or you want to see the order- $m$ formula applied to a pole of order $3$ , ask the AI tutor for a few more practice problems.

How do I tell a pole of order 2 from two separate simple poles at the same point?

You cannot have two separate simple poles at the same point — a single point has exactly one Laurent series about it, and that series has one, unambiguous principal part. If the principal part's most negative surviving term is $c_{-2}/(z-z_0)^2$ , that is a single pole of order $2$ , full stop. The confusion usually comes from a denominator like $(z-z_0)^2$ arising as a product of two distinct linear factors that happen to be equal, which is exactly what "order $2$ " already encodes — there is no separate case to worry about.

Why does the residue only care about the c minus 1 term and not the whole principal part?

Because that is exactly what survives contour integration. Integrating $(z-z_0)^n$ around a small closed loop about $z_0$ gives $0$ for every integer $n$ except $n=-1$ , where it gives $2\pi i$ . When you integrate the full Laurent series term by term around such a loop, every term vanishes except the one multiplying $c_{-1}$ . So no matter how complicated the principal part is, only its $1/z$ coefficient ever shows up in a contour integral — which is precisely why it earns the name "residue," what is left over after everything else cancels.

Do I always need the full order-m formula, or is there a shortcut for simple poles written as a fraction?

For a simple pole of $f = p/q$ with $q(z_0)=0$ and $q'(z_0)\ne 0$ , you can skip the limit-of-a-product form entirely and use $\operatorname{Res}_{z=z_0}f(z) = p(z_0)/q'(z_0)$ , which follows from the order- $1$ case of the formula above by recognizing $\lim_{z\to z_0}(z-z_0)/q(z)$ as $1/q'(z_0)$ from the definition of the derivative. It is faster whenever $q$ factors are awkward to isolate by hand, but it only applies at simple poles — for order $2$ or higher you need the full derivative formula.

Zeros, singularities, and residues

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads