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First think of c as being represented by z zt the

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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...
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