1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

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814 CHAPTER 8 8 RESIDUE THEORY

provided R 2 R~. The parametrization of CR leads to the equation


ldzl = R d(J and ie"I = e- 11 = e-Rsin8. (8-17)

Using t he trigonometric identity sin (n - 8) = sinB and Equations (8-17), we
ex press the integral on the right s ide of Inequality {8- 16 ) as


( : iei" I ldzl = : (" e-Rsin8 R d8 = 2e fr; e -Rsin8 R dB.

l cn 1T 1f lo 1f lo


(8-18)

On the interval 0 :::; (J :::; ~ we can use t he inequality

We combine this inequality with Inequality (8- 1 6) and Equation (8-18) to con-
clude that, for R 2 R~,

I

r exp(iz)P(z)dzl ~ 2e r~ e -~R· R d8
l cn Q (z) 1T lo

= - ee- .-->RB^1 8=~ = e (1 - e-R) < e.
O=O
Because e > 0 is arbitrary, our proof is complete.


  • We now turn to t he proof of our main t heorem.

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