Complex Functions Examples c-6 - Calculus of Residues by Mejlbro L.

By Mejlbro L.

This can be the 6th textbook you could obtain at no cost containing examples from the speculation of advanced features. during this quantity we will ponder the foundations of calculations or residues, either in finite singularities and in ∞. the speculation seriously is dependent upon the Laurent sequence from the 5th ebook during this sequence. The purposes of the calculus of residues are given within the 7th ebook.

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If c2 = 0, then the singularity at 0 becomes removable. According to the above we have a−1 = 0, so res(f ; 0) = 0 for any such solution, and we have answered all questions with the exception of determining the recursion formula, which does not give sense any more. com 53 Complex Funktions Examples c-6 Line integrals computed by means of residues Second method. e. the series method. By inserting a formal Laurent series f (z) = an z n and its derivatives f (z) = n an z n−1 and f (z) = n(n − 1)an z n−2 , we get 0 = = = z 2 − z f (z) + (5z − 4)f (z) + 3 f (z) n(n−1)an z n − n(n−1)an z n−1 + n2 −n+5n+3 an z n − 5nan z n − 4nan z n−1 + 3an z n n(n+3)an z n−1 = (n+1)(n+3)an z n − n(n+3)an z n−1 = {(n+1)(n+3)an −(n+1)(n+4)an+1 } z n = (n+1) {(n+3)an − (n+4)an+1 } z n .

NNE Pharmaplan offers me freedom with responsibility as well as the opportunity to plan my own time. com NNE Pharmaplan is the world’s leading engineering and consultancy company focused exclusively on the pharma and biotech industries. NNE Pharmaplan is a company in the Novo Group. 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 1 lies inside the path of integration |z| = 1, and it is a pole of second order, so 2 ⎫ ⎧ ⎛ ⎞ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎜ ⎟ ⎨ z z z e e d 1 ⎟ 2πi e ⎜ dz = 2πi · res ; = lim ⎜ ⎟ 2 2 2 2⎠ 1!

E. such that the imaginary unit does not occur. Since F (z) = 1 (z 2 + 1) 2 has the two double poles ±i, we shall only find f (t) = res ezt (z 2 + 1) 2 ;i + res ezt (z 2 + 1) 2 ; −i . We get by Rule I, res ezt (z 2 + 1) 2 ;i ezt (z + i)2 t ezt 2 ezt − (z + i)2 (z + i)3 = d 1 lim z→i dz 1! com 48 Complex Funktions Examples c-6 Line integrals computed by means of residues and res ezt (z 2 + 1) 2 ; −i = d 1 lim 1! z→−i dz −it = ezt (z − i)2 = lim z→−i t ezt 2 ezt − 2 (z − i) (z − i)3 −it te 2e 1 i − = − t e−it + eit , 2 3 (−2i) (−2i) 4 4 hence by insertion into (8), i 1 i 1 1 f (t) = − t eit − eit − t e−it + e−it = − t 4 4 4 2 4 1 1 = − t cos t + sin t.

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