proof of radius of convergence of a complex function


Without loss of generality, it may be assumed that z0=0.

Let cn denote the coefficient of the n-th term in the Taylor seriesMathworldPlanetmath of f about 0. Let r be a real number such that 0<r<R. Then cn may be expressed as an integral using the Cauchy integral formulaPlanetmathPlanetmath.

cn=12πi|z|=rf(z)zn+1𝑑z=12πrn-π+πe-nθf(reiθ)𝑑θ

Since f is analytic, it is also continuousMathworldPlanetmath. Since a continuous function on a compact set is bounded, |f|<B for some constant B>0 on the circle |z|=r. Hence, we have

|cn|=12πrn|-π+πe-nθf(reiθ)𝑑θ|12πrn-π+π|e-nθf(reiθ)|𝑑θ12πrn-π+πB𝑑θ=Brn

Consequently, cnnBn/r. Since limnBn=1, the radius of convergenceMathworldPlanetmath must be greater than or equal to r. Since this is true for all r<R, it follows that the radius of convergence is greater than or equal to R.

Title proof of radius of convergence of a complex function
Canonical name ProofOfRadiusOfConvergenceOfAComplexFunction
Date of creation 2013-03-22 14:40:35
Last modified on 2013-03-22 14:40:35
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 9
Author rspuzio (6075)
Entry type Proof
Classification msc 30B10