By Leif Mejlbro

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**Additional resources for Complex Functions Examples c-6 - Calculus of Residues**

**Example text**

Lim = − lim 1 z→∞ z→∞ z 2 (z − 1) 1− z z2 + z + 1 ;∞ z 2 (z − 1) (b) Since we have only a ﬁnite number of singularities in C, and since the sum of the residues is zero, we get res(f ; ∞) = −res(f ; 0) − res(f ; 3i) − res(f ; −3i). Here z = 0 is a double pole, so res(f ; 0) = ez z2 + 9 d dz 1 1! ez + z {· · · } z2 + 9 = z=0 = z=0 1 . 9 Since z = ±3i are simple poles, we get res(f ; 3i) = e3i 1 e3i e3i = = − · , (3i)2 (3i + 3i) −27 · 2i 27 2i res(f ; −3i) = e−3i e−3i 1 e−3i = = · , (−3i)2 (−3i − 3i) 27 · 2i 27 2i hence 1 1 res(f ; ∞) = − − 9 27 − e3i − e−3i 2i 1 1 sin 3.

Alternatively we apply Rule I. Then res f ; − π 2 = 1 z− limπ z→− 2 = − 1 · π π 2 · z + π2 1 = π cos z −2 − 1 limz→− π2 d cos z dz 1 cos z − cos − π2 limz→− π2 z − − π2 1 1 1 =− · =− . π [− sin z]z=− π2 π π 2 · π . Here we shall demonstrate a seldom application of Rule III, where 2 π B(z) = z − cos z. com 38 Complex Funktions Examples c-6 Line integrals computed by means of residues hence res f ; π 2 = 6A B − 2AB 6 · 0 · (−2) − 2 · 1 · 0 = 0. 3 · (−2)2 = 2 3 (B ) Alternatively (and more diﬃcult) we use Rule I and l’Hospital’s rule (or possibly o-technique) with π π π z− z− z− 2 2 2 g(z) = = =− π π , cos z sin −z sin z − 2 2 hence π g 2 π 2 = − limπ π z→ 2 sin z − 2 z− = −1, and π g(z) − g 2 π z− 2 z − π2 π +1 z − + cos z T (z) cos z 2 = = .

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 · .