Laurent series and isolated singularities
Classify punctured-neighborhood behavior using negative powers and keep each expansion tied to its annulus.
On this page
The question for today
What you’ll understand
- Construct Laurent expansions on specified annuli.
- Classify removable singularities, poles, and essential singularities.
- Explain why a zero residue does not determine a singularity's type.
Taylor series describe holomorphic behavior near an ordinary point. Laurent series add negative powers so that a function holomorphic on a punctured neighborhood can reveal how it fails at the missing center. The pattern of those negative powers distinguishes a removable gap, a pole, and an essential singularity. The expansion is always attached to an annulus of analyticity. A different annulus can have different coefficients, so extracting a coefficient without stating where its series converges can produce a misleading local conclusion.
An annulus supplies two sides of the expansion
If f is holomorphic on an annulus , it has a unique Laurent series there, converging locally uniformly on that annulus. The nonnegative powers form a Taylor-like part and the negative powers form the principal part. Coefficients are given by integrals around any positively oriented centered circle lying within the annulus. Cauchy's theorem makes the coefficient integral independent of the chosen radius while the circle stays inside that same analytic annulus.
The inner and outer radii matter. Expanding a rational factor by a geometric series requires the chosen ratio to have modulus less than one. A ratio appropriate inside a pole radius is invalid outside it, where rewriting in inverse powers yields another series. Neither algebraic identity overrides the convergence requirement. At the boundary circles, ordinary series convergence must be examined separately; the Laurent theorem guarantees the open annulus, not automatic convergence on its edges.
The principal part classifies an isolated point
An isolated singularity at a means f is holomorphic on some full punctured disk . If every negative coefficient vanishes, the missing value can be filled by the constant coefficient and the singularity is removable. If only finitely many negative coefficients occur and the most negative is −m, the point is a pole of order m. Then multiplication by produces a holomorphic function nonzero at a. Infinitely many nonzero negative coefficients define an essential singularity.
The coefficient of is the residue. It is crucial for contour integration, but it does not by itself classify the singularity. A second-order pole can have zero residue, and an essential singularity can also have zero residue. For example, the exponential of 1/z² has infinitely many negative even powers and no negative first power. A branch point such as the origin for a continued logarithm is not an isolated singularity of one single-valued holomorphic function on a full punctured disk.
Do not confuse an outer coefficient with a local residue
A Laurent coefficient computed on an annulus that surrounds several excluded points records the behavior on that annulus. Its negative first-power coefficient determines the integral on its centered circles, but it need not equal the local residue at the center. The local residue uses sufficiently small circles enclosing only the isolated central singularity. Crossing another pole changes the available annulus and can change the centered Laurent coefficients, even though the rational function itself is unchanged.
This distinction is useful when preparing a residue calculation. First list singularities and their distances from the chosen center. Then divide the plane into maximal circular annuli avoiding them, expand each factor using a convergent geometric ratio, and check the first few terms against the original expression. Classification of the central singularity uses the innermost punctured disk. Integrals on larger circles can be understood later as sums of local residues rather than silently assigning an outer coefficient to one point.
Work it through
One rational function, two Laurent annuli
Expand about zero on and on . Find the local residue and explain the different negative first-power coefficients.
For the inner annulus, write . Expanding the geometric series gives , valid only for with zero excluded.
The central point has a simple pole with local residue . For the outer annulus instead write , valid for .
The outer series has no z⁻¹ term. Its circles enclose both poles 0 and 2; the second pole's residue is , canceling the first. An outer centered coefficient of zero therefore does not change the local residue at zero.
The local residue is −1/2. The outer annulus has negative first-power coefficient zero because its contour surrounds a different collection of singularities.
The living notebook
Explore the model
Interactive lab · vectors
Inspect reciprocal behavior away from the missing point
Pause and Reset; set Angle=45 degrees,Magnitude=2. Predict z=√2+i√2. In a fresh comparison Reset and set Angle=−45 degrees,Magnitude=0.5: the vector is 1/z, with conjugated phase and reciprocal modulus. This illustrates the negative first-power factor around a nonzero point. It does not classify arbitrary singularities, compute Laurent coefficients, or allow a reciprocal value at zero.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forNegative powers invert radial scale and phase, while Laurent classification requires the complete principal part on a punctured disk.
Pause & reason
Check your understanding
Your turn
Try it yourself
Classify the singularity at zero of and calculate its residue. Is zero residue enough to remove it?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Expand sine far enough to see the first nonzero negative power after subtracting z and dividing by z⁵.
Compare with the solution
The numerator is . Hence . The first term is a nonzero negative second power, so zero is a pole of order 2. There is no z⁻¹ term, giving residue zero. Filling one value cannot remove the nonzero second-order principal part.
Keep the idea
Take it with you
- Every Laurent expansion belongs to a specified annulus.
- The whole principal part distinguishes removable, pole, and essential behavior.
- Local residues must not be replaced by coefficients from an outer annulus enclosing other poles.
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