Laurent series on annuli

Try

The same rational function can have two valid series about the same center. For 1/(z−1), the disk |z|<1 gives −1−z−z²−⋯. Outside |z|>1, the expansion is z^(−1)+z^(−2)+⋯. The pole on the separating circle prevents these from being one shared Taylor expansion.

Laurent series allow both positive and negative integer powers. Their natural domain is an annulus, described by two radii around a center. Specifying only the center leaves essential convergence information missing.

Predict

Predict the coefficient of z^(−1) in the two expansions above. It is zero inside and one outside. This does not violate uniqueness: uniqueness holds on a specified annulus.

Also predict which terms survive integration around a circle centered at zero. The earlier integer-power computation says only exponent minus one survives. This will make Laurent coefficients directly observable through contour integrals.

Experiment

Expand 1/[z(z−2)] in two regions.

Annulus	First terms	Coefficient of z^(−1)
0<∣z∣<2	−1/(2z)−1/4−z/8−⋯	−1/2
∣z∣>2	z^(−2)+2z^(−3)+⋯	0

Choose a test point well inside each region and check the geometric ratio has modulus below one. A formally attractive series with ratio above one is not a valid representation.

Understand

Let f be holomorphic on r<|z−a|<R. Choose an inner circle and outer circle within the annulus, with the evaluation point between them. The Cauchy formula for the region between those circles expresses f(z) as the outer kernel integral minus the inner kernel integral.

On the outer circle expand 1/(ζ−z) in nonnegative powers of (z−a)/(ζ−a). On the inner circle use the opposite ratio: 1/(ζ−z)=−Σ(ζ−a)ⁿ/(z−a)^(n+1). Its minus sign cancels the inner-boundary subtraction. The two expansions produce a doubly infinite series Σcₙ(z−a)ⁿ, with n ranging over all integers. Each part converges uniformly on compact subannuli, so integration and summation are justified there.

For every integer k, multiply the series by (z−a)^(−k−1) and integrate around an intermediate positive circle. All powers integrate to zero except exponent minus one. Hence c_k=(1/(2πi))∫f(z)/(z−a)^(k+1)dz. Cauchy’s theorem on annular strips shows this coefficient does not depend on the chosen circle within the annulus. The same formula proves uniqueness.

The sum of negative-power terms is the principal part. It may contain finitely many nonzero terms, infinitely many, or none. Those three possibilities will classify isolated singularities when the inner radius is zero.

Apply

Derive the two table rows yourself by factoring the denominator differently. For the inner region, extract −2z and expand in z/2. For the outer region, extract z² and expand in 2/z. Explain why using the inner series at z=3 fails.

Repair this claim: “The z^(−1) coefficient is always the residue at the center.” It is the local residue when the series holds on a punctured disk around that center. An exterior annulus may enclose other singularities, and its coefficient records their combined contour contribution instead.

Continue

The next lesson interprets principal parts near isolated missing points. Optional depth: derive a Laurent series on 1<|z|<3 for 1/[(z−1)(z−3)] by partial fractions, choosing an exterior expansion for one factor and an interior expansion for the other. State the overlap before combining them.