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# This gives o fz dz o zz zi dz si si diam

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Unformatted text preview: tself. Then for any œ&gt;0, R can be covered by a finite number of (partial) squares so that each (partial) square Si contains a fixed point zi such that for each z é Si , one has Ô Ô Ôf(z) - f(zi) Ô - f'(zi)Ô &lt; œ Ô Ô z - zi Ô Remarks on why that is true: Certainly, we can cover the region R by infinitely many such squares, and the result now follows by the fact that the region R is compact. Now do a little algebra to write f(z) = f(zi) + f'(zi)(z-zi) + ©(z)(z-zi) where k is the expression inside the absolute values above. One therefore has Û Û Û Û Ù Ù Ù Ù O f(z) dz = O f(zi) dz + O f'(zi)(z-zi) dz + O ©(z)(z-zi) dz ı ı ı ı Si Si Si Si 1 Don't bother with the textbook's proof -- they only prove a special case by citing Green's theorem, which few instructors have time to prove in calc 2 anyway.. 14 Û Û Ù Ù However, f(zi) is a constant, and O 1 dz and O z dz = 0 for any closed contour, since ı ı Si Si the functions f(z) = 1 and f(z) = z posses antiderivatives. . Therefore, we are left with Û Û Ù Ù O f(z) dz = O ©(z)(z-zi) dz ı ı Si Si Now |©(z)| &lt;...
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