Removable singularities, poles and essential singularities

Try

The three functions sin z/z, 1/z² and exp(1/z) all fail to be defined at zero as written, but their missing points are fundamentally different. The first can be repaired by assigning the value one. The second grows like a finite negative power. The third contains infinitely many negative powers and has no pole order.

An isolated singularity means the function is holomorphic on some punctured disk around the point. This lesson does not classify nonisolated singular sets or branch points with no full punctured-disk branch.

Predict

Predict whether boundedness near a missing point is enough to repair it holomorphically. For a holomorphic function on a punctured disk, it is. Merely being bounded as an arbitrary function would not be enough.

Predict also whether an essential singularity must attain every complex value near the point. The exponential never equals zero, so exp(1/z) immediately warns against that stronger claim.

Experiment

Classify using the Laurent principal part.

Function near zero	Principal part	Type
sin z/z	None	Removable
(1+z)/z³	z^(−3)+z^(−2)	Pole of order 3
exp(1/z)	Infinitely many negative terms	Essential

The pole order is the largest negative exponent magnitude with nonzero coefficient. The residue, introduced next, is only the coefficient of exponent minus one.

Understand

A removable singularity has no negative Laurent coefficients, so its nonnegative series defines a holomorphic extension. Conversely an extension has an ordinary Taylor series and therefore no principal part.

Here is the boundedness criterion with proof. Suppose |f|≤M on a punctured disk about a. For n≥1, the coefficient c_(−n) is the integral of f(z)(z−a)^(n−1) divided by 2πi on any sufficiently small positive circle. ML gives |c_(−n)|≤Mrⁿ. Letting r tend to zero makes every such coefficient vanish. Thus f extends holomorphically.

A pole of order m has f(z)=(z−a)^(−m)g(z), where g is holomorphic and g(a)≠0. Its reciprocal extends with a zero of order m. Conversely a zero of order m in the reciprocal gives precisely this factorization. In particular a pole’s modulus tends to infinity uniformly as the point is approached.

An essential singularity is neither removable nor a pole. Casorati–Weierstrass states that its image on every punctured neighborhood is dense in C. To prove this, suppose an open disk about a target value w is omitted. Then 1/(f−w) is bounded and extends across the point. If its extension value is nonzero, f extends. If it is zero, its finite zero order makes f have a pole. The reciprocal cannot vanish identically because it is nonzero on the punctured neighborhood. Both alternatives contradict essentiality.

Apply

For exp(1/z), choose positive real z tending to zero to obtain unbounded values, and negative real z tending to zero to obtain values tending to zero. These two paths already exclude both removability and a pole. The Laurent series gives the complete classification independently.

Diagnose the invalid deduction “Dense image means every value is attained.” A dense set can omit individual points. Casorati proves meeting every target disk, whereas Picard’s much stronger theorem is not proved by the reciprocal argument here.

Continue

Classify first, then calculate. This prevents confusing a zero residue with a removable point: 1/z² has residue zero but a genuine pole. Optional depth: prove that a finite limit at an isolated singularity yields the removable extension, and compare that criterion with boundedness, which does not initially specify the limiting value.