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Power Series and Analyticity in the Complex Plane

Expand an analytic function as a power series on a disc, and read the radius of convergence off the nearest singularity.

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What a learner can do afterwards

  • Expand a function about a point and state the disc on which the series is valid
  • Explain a real series' radius of convergence by pointing at a complex singularity
  • Show that a zero of a function analytic on a connected open domain is isolated unless the function is identically zero

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A power series is a sum over n ≥ 0 of c_n (z - a)^n with centre a. At z = a it always converges to c_0, including when R = 0. For R > 0 it converges on the open disk |z - a| < R, diverges outside, and each boundary point is tested individually. Inside the disk the sum is holomorphic. The radius of the Taylor expansion of an analytic function is the distance to the nearest genuine nonremovable obstruction, not an arbitrary chosen boundary or a removable hole. Entire functions have infinite radius. Analytic means locally represented by a convergent power series.

sum c_n (z-a)^nconverges at centreopen disk |z-a| < Ranalytic = local series
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Worked geometric series. The infinite sum 1 + z + z² + ... equals 1/(1 - z) exactly when |z| < 1. The partial sum through z^N is (1 - z^(N+1))/(1 - z). The exact tail. The discarded tail after z^N is z^(N+1)/(1 - z). For |z| ≤ r < 1 the tail is bounded in modulus by r^(N+1)/(1 - r), an absolute bound rather than a signed error. On |z| = 1 every geometric term has modulus 1, so this particular boundary fails the term test at every point. Other series may converge at some boundary points.

tail = z^(N+1)/(1-z)|tail| ≤ r^(N+1)/(1-r)|z|=1 fails here

From Cauchy to Taylor. Fix the circle |w - a| = r traversed once counterclockwise, with f holomorphic on a neighbourhood of its closed disk and z strictly inside. Expand 1/(w - z) as the sum over n ≥ 0 of (z - a)^n/(w - a)^(n+1). On the circle the ratio |z - a|/r is below 1 uniformly, so the geometric series converges uniformly and sum and integral may be exchanged. (1/(2πi)) times the integral of f(w)/(w - a)^(n+1) dw equals f^(n)(a)/n!. So f equals its Taylor series on the disk: holomorphic implies analytic. Convergent power series differentiate term by term within their disk.

Worked shifted expansion. For f(z) = 1/(1 - z) about a = 2, rewrite 1 - z as -(1 + (z - 2)) and expand geometrically: f(z) is the sum of (-1)^(n+1) (z - 2)^n for |z - 2| < 1, with alternating signs and radius 1 from the pole at 1. Contrast the real function 1/(1 + x²): no real blowup at x = 1, yet complex denominator zeros at ±i sit one unit from 0, capping the radius. Smooth real functions need not equal their Taylor series.

sum (-1)^(n+1)(z-2)^n|z-2| < 1poles ±i at distance 1

Operations on series. Work with same-centre series. Term by term differentiation and integration preserve the exact radius. Coefficient-wise sums and Cauchy products are valid on the common disk, and cancellation can increase the radius: the coefficient product (1 - z)(1 + z + z² + ...) gives the constant series 1 with infinite radius, extending the coefficient series rather than evaluating the geometric factor outside its disk. The ratio formula R = 1/L needs the limit L of |c_(n+1)/c_n| to exist with quotients eventually defined, with L = 0 giving infinite and L infinite giving 0. Sparse zero coefficients can make the ratio unavailable.

termwise: exact radiusproducts: can growR = 1/L if defined

Local zero factorization. Let f be analytic on an open domain with f(a) = 0. If some Taylor coefficient is nonzero, let m ≥ 1 be the first such order. Then f(z) = (z - a)^m g(z) with g analytic and g(a) nonzero. By continuity g stays nonzero on a small disk around a, so a is the only zero there: the zero is isolated of order m. Example: z³ - z = z(z - 1)(z + 1) has simple zeros at 0, 1 and -1. If every coefficient vanishes, the Taylor series makes f locally zero.

Global identity. Let f be analytic on a connected open domain D, and let S be the set of points where f vanishes on a neighbourhood. S is open by definition. If a is an interior limit point of S, continuity of every derivative gives f^(n)(a) = 0 for all n, so the Taylor series makes a belong to S; hence S is closed relative to D. Connectedness forces S empty or all of D. Distinct zeros accumulating at an interior point rule out a first nonzero coefficient by the local factorization, forcing f identically zero. Boundary points, infinity, and other components escape this conclusion.

The nearest genuine obstruction sets the Taylor radius, coefficients come from derivatives, and zeros of a nonzero analytic function stand apart.

All nine extra tasks and four readings are original authorship created for this topic.

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Power Series and Analyticity in the Complex Plane · Mathematics, ages 20 to 21 · LightMySky