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Unformatted text preview: tself. Then for any œ>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 < 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)(zzi) + ©(z)(zzi)
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)(zzi) dz + O ©(z)(zzi) 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)(zzi) dz
ı
ı
Si Si Now ©(z) <...
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 Fall '03
 StefanWaner
 Math, Algebra, Geometry, Complex Numbers

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