# proof of radius of convergence of a complex function

Without loss of generality, it may be assumed that ${z}_{0}=0$.

Let ${c}_{n}$ denote the coefficient of the $n$-th term in the Taylor series^{} of $f$ about $0$. Let $r$ be a real number such that $$. Then ${c}_{n}$ may be expressed as an integral using the Cauchy integral formula^{}.

$${c}_{n}=\frac{1}{2\pi i}{\oint}_{|z|=r}\frac{f(z)}{{z}^{n+1}}\mathit{d}z=\frac{1}{2\pi {r}^{n}}{\int}_{-\pi}^{+\pi}{e}^{-n\theta}f(r{e}^{i\theta})\mathit{d}\theta $$ |

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

$$|{c}_{n}|=\frac{1}{2\pi {r}^{n}}\left|{\int}_{-\pi}^{+\pi}{e}^{-n\theta}f(r{e}^{i\theta})\mathit{d}\theta \right|\le \frac{1}{2\pi {r}^{n}}{\int}_{-\pi}^{+\pi}|{e}^{-n\theta}f(r{e}^{i\theta})|\mathit{d}\theta \le \frac{1}{2\pi {r}^{n}}{\int}_{-\pi}^{+\pi}B\mathit{d}\theta =\frac{B}{{r}^{n}}$$ |

Consequently, $\sqrt[n]{{c}_{n}}\le \sqrt[n]{B}/r$. Since ${lim}_{n\to \mathrm{\infty}}\sqrt[n]{B}=1$, the radius of convergence^{} must be greater than or equal to $r$. Since this is true for all $$, 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 |