By Mejlbro L.

This is often the 6th textbook you could obtain at no cost containing examples from the speculation of complicated services. during this quantity we will ponder the principles of calculations or residues, either in finite singularities and in ∞. the speculation seriously depends upon the Laurent sequence from the 5th e-book during this sequence. The functions of the calculus of residues are given within the 7th ebook.

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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 · . i (c) We get from 1 1 ≤ m 2 2 |a + bz | (bR − a) for |z| = R > a , b the estimate lim R→+∞ CR dz 1 m ≤ lim m · πR = 0, R→+∞ (bR2 − a) (a + bz 2 ) proving that lim R→+∞ CR dz m = 0.

Com 37 Complex Funktions Examples c-6 Line integrals computed by means of residues 2 1 –2 –1 1 2 –1 –2 Figure 3: The curve |z| = 2 with the two poles insider. Determination of res f ; − A(z) = 1 z− π π . The pole z = − is simple. g. 2 2 and π 2 B(z) = cos z. Then π res f ; − 2 π 2 π − 2 A − = B = 1 π π π · − sin − − − 2 2 2 1 =− . π 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 =− · =− .

Z n z z z z 2 z3 for z = 0. 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.

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