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Additional resources for Complex Functions Examples c-6 - Calculus of Residues
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1 z4 (a) Here, z = i is a pole of at most second order inside |z| = 2, so |z|=2 z 3 − 3z + 1 d 2πi z 3 − 3z + 1 = 2πi lim 3z 2 − 3 = −12πi. dz = lim z→i (z − i)2 1! z→i dz (b) Here, z = 0 is a pole of at most seventh order inside |z − 1| = 2, so |z−1|=2 cos z d6 2πi 2πi 2πi πi dz = lim cos z = lim (− cos z) = − =− . 7 6 z→0 z→0 dz z 6! 6! 6! 360 Alternatively, it follows by a series expansion for z = 0 that z4 z6 1 z2 cos z + − + ··· = · 1 − 2! 4! 6! z7 z7 = 1 1 1 1 1 1 1 − + − + ··· , 7 5 3 z 2 z 24 z 720 z and since res(f ; 0) = a−1 , we get |z−1|=2 cos z 2πi πi dz = 2πi · res(f ; 0) = − =− .
In particular, res(f ; 0) = a−1 = 0 for every convergent series solution. The case n = −1 has already been treated above. If n ∈ N0 , then it follows from (12) that (n + 4)an+1 = (n + 3)an = · · · = (0 + 3)a0 = 3a0 , hence an = 3 a0 n+3 for n ∈ N0 , and we have found all coefficients. Summing up, the formal Laurent series solutions are given by +∞ 3 a−3 zn, (13) f (z) = 3 + a0 z n + 3 n=0 and it follows that the domain of convergence in general is 0 < |z| < 1 for a−3 = 0 and a0 = 0. Please click the advert Student Discounts + Student Events + Money Saving Advice = Happy Days!
The series expansion of cos z from z = is given by 2 2 N (z) = z − cos z = − z − π 1 π + z− 2 3! 2 3 z− +o π 2 3 , hence 1 π z− 6 2 3 3 +o z− π 2 +o z− π 2 2 T (z) 1 π =− z− +o N (z) 6 2 z− π 2 1 T (z) = N (z) = − z − π 2 2 , , and we conclude that , and therefore, π res f ; 2 1 = g 1! π 2 π g(z) − g 2 = limπ π z→ 2 z− 2 = limπ z→ 2 T (z) = 0. com 39 Complex Funktions Examples c-6 Line integrals computed by means of residues Alternatively we apply l’Hospital’s rule recursively, since T (z) = z − π2 + cos z, N (z) = z − π2 cos z, T N T (z) = 1 − sin z, N (z) = cos z − z − T N π 2 T (z) = − cos z, N (z) = −2 sin z − z − sin z, π 2 cos z, T N π 2 π 2 π 2 π 2 π 2 π 2 = 0, = 0, = 0, = 0, = 0, = −2, and we conclude again that limπ z→ 2 T (z) 0 = = 0.