VII.1.2 Calculate the residue at each singularity in the complex plane of the following functions. (a) e (b) tanz (c) (22+1)2 (d) z2z Solution (a) The function e/ has a singularity at z = 0, by Laurent expansion Thus by definition (b) 1 1 el/z =1++ + 2!22 Res [e,0] = 1. The function tan z = sinz has isolated singularities at z = +n, - < COS 2 n, they are simple poles. By rule 3, sin z sin z Res [tanz, /2+n] = Res /2 + n =-1. COS 2 sin z z=/2+n (c) The function (241)2 has isolated singularities at z = i, they are double poles. By rule 2, d z Res = = dz (z+i) z=i Z d Z Res = (x + 1)21 dz (z-i) z=-i (z+i) - 2z (z+i) (z+i)4 (z - i) - 2z (z - i) (z-i)4 = 0, z=i = 0.