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By Leif Mejlbro
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M − 1)! com 44 Complex Funktions Examples c-6 Line integrals computed by means of residues it follows by insertion that f (z) 1 = a b bm z−i +∞ × (−1)n n=0 m 1 · m a b 2i (m+n−1)! (m−1)! × 1 a b 2i +∞ (−1)m (m+n−1)! (m−1)! n=0 = 1 2i z−i m+n a b +∞ 2m+p−1)! (m−1)! p=−m = +∞ ap z − i = p=−m a b a b z−i n n 1 a b 2i n−m a b 2m+p z−i a b p p , which is the Laurent series expansion of f (z) in the set 0< z−i of centrum i a <2 b a , b a and radius R = 2 b a . e. a−1 = 1 (2m − 2)! (m − 1)! (−1)m−1 2i a b 2m−1 = 1 bm 2m − 2 m−1 1 · 2 a b 2m−1 1 · .
1 , then z sin z 1 1 = = f (z), (−z) · sin(−z) z sin z so the integrand is an even function. Then by (1) we have in particular a−1 = 0, because −1 is an odd index. Then |z|=1 1 dz = 2πi res z sin z 1 ;0 z sin z = 2πi a−1 = 0. 2 Find the value of the line integral |z|=2 ez dz. com 33 Complex Funktions Examples c-6 Line integrals computed by means of residues It is not a good idea in this case to use the traditional method of inserting a parametric description and then compute. Note instead that we have inside the curve |z| = 2 (seen in its positive direction) the two isolated singularities z = 0 and z = 1, hence by Cauchy’s residue theorem, |z|=2 ez dz = 2πi res z(z − 1)2 ez ; 0 + res z(z − 1)2 ez ;1 z(z − 1)2 .