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Alternatively we use that we here also have res(f ; 1) = −res(f ; ∞). It is actually possible directly to find res(f ; ∞), but again the computations are rather difficult. Alternatively we expand g(z) = e2z − ez+1 ud fra z = 1. as a series. 3e2 , g (3) (1) = 7e2 , g (4) (1) = 15e2 , so the Laurent series expansion becomes f (z) = g(z) 1 = (z − 1)5 (z − 1)5 0+ 3e2 7e2 15e2 e2 (z − 1) + (z − 1)2 + (z − 1)3 + (z − 1)4 + · · · 1! 2! 3! 4! From here we get res(f ; 1) = a−1 = 15e2 5e2 15e2 = = , 4! 24 8 hence by insertion, |z|=2 e2z − ez+1 5πe2 · i.

N=0 z ∈ C \ {0}, that res(f ; 0) = a−1 = 1 1 = , 2! 2 hence |z|=1 ez dz = 2πi · res(f ; 0) = 2πi · a−1 = πi. 7 Compute |z|=2 z dz. 4 by Rule II. We shall here show that it is much easier to use Rule IV instead, because |z|=2 ∗ ∗ |z|=2 z dz = −2πi · res z4 − 1 z ;∞ z4 − 1 z2 = 0, z→∞ z 4 − 1 = 2πi · lim denotes that we have changed the direction of the path of integration ∗ C · · · dz = − C · · · dz. 8 Prove that ez |z|=1 z2 3 +z− 4 2 dz = 0. The poles of the integrand are given by 1 z=− ± 2 1 1 3 + = − ± 1, 4 4 2 thus z1 = 1 2 Only z1 = |z|=1 and 3 z2 = − .

2 Find the residues at ∞ of the following functions: (a) 1 , z (1 − z 2 ) (b) (a) The function f (z) = res z4 (z 2 + 1) (c) z 2n , (1 + z)n 1 ;∞ z (1 − z 2 ) = 0. ⎛ 1 ;∞ z (1−z 2 ) n ∈ N. 1 has a zero of order 3 at ∞, so z (1 − z 2 ) Alternatively, res 2, ⎞ ⎜1 = −res ⎜ ⎝ z2 · 1 z 1 1− 1 z2 ⎟ ; 0⎟ ⎠ = −res z z 2 −1 ;0 = 0. (b) The Laurent series expansion of the function f (z) = z4 (z 2 + 1) 2 only contains even powers of z, so a−1 = 0, and thus res z4 (z 2 + 1) 2 ;∞ = 0. (c) It follows by the rules of computation, ⎛ ⎞ 2n 1 ⎜1 ⎟ z 2n z ⎜ ⎟ res ; ∞ = −res ⎜ 2 · n ; 0⎟ = −res n (1 + z) ⎝z ⎠ 1 1+ z 1 1 · ;0 z n+2 (z + 1)n 1 1 dn+1 n(n + 1) · · · (n + n + 1 − 1) 1 lim n+1 =− lim (−1)n+1 · (z + 1)n (n + 1)!

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