First think of c as being represented by z zt the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Û n!M f(z) Ù ÔO dzÔ ≤ 2π n+1 2πr = n 2π Ô ı (z - z )n+1 Ô r r 0 ÔC Ô as required. Corollary 5.5 (Louville's Theorem) Entire bounded functions are constant. Proof: S’pose that f is bounded on the entire complex plane, so that |f(z)| ≤ K for some constant K. We now use the case n = 1 of the above theorem, giving K |f'(z0)| ≤ r where r is the radius of an arbitrary circle with center z0. Since r is arbitrarily large, it must be the case that f'(z0) = 0. Since this is true for every z0 é C , it must be the case I that f(z) = constant. (If f'(z) = 0, then the partial derivatives of u and v must all vanish, and so u and v are constant.) n! 2π Û M Ù O ı |z - z |n+1 dz 0 C The integrand is now constant, since |z - z0| = r, the radius of the circles. Therefore, the integral on the right boils down to n!M Û Ù n O dz 2πr ı C ≤ Corollary 3 (Fundamental Theorem of Algebra) Every polynomial function of a complex variable has at least one zero. 1 is entire. But it is also...
View Full Document

This document was uploaded on 03/20/2014 for the course MATH 144 at Hofstra University.

Ask a homework question - tutors are online