Math 412 Group Work #14 Spring 2009
Problem 1. Write down the higher-order version of the Cauchy Integral
Formula. Be sure to include all necessary hypotheses.
SOLUTION: This is Theorem 6.12 (p. 237).
Problem 2. Evaluate:
(a):
ZC
eiz
(z−1)3dz,
where C=C+
3(0).
SOLUTION: We apply the Cauchy Integral Formula (higher-order version) with
z0= 1, n= 2, and f(z) = eiz . We have f0(z) = ieiz and f00(z) = −eiz. Then
f00(z0) = −ei=−[cos 1 + isin 1]. Therefore, the answer is
ZC
eiz
(z−1)3dz =−[cos 1 + isin 1] ·2πi/2! = −iπ[cos 1 + isin 1] = πsin 1 −i(πcos 1).
(b):
ZC
cos z
(z+π)5dz,
where C=C+
4(0).
SOLUTION: We apply the Cauchy Integral Formula (higher-order version) with
z0=−π,n= 4, and f(z) = cos z. We have f0000(z) = cos z, so
ZC
cos z
(z+π)5dz =2πi
4! cos(−π) = −iπ
12.
(c):
ZC
ezcos z
z2(z−1)2dz,
where C=C−
1/2(3/4).
SOLUTION: Let us first determine the contour integral for −C; we’ll insert a minus
sign at the end to correct for this. Note that
f(z) = ezcos z
z2
