By Mejlbro L.

This can be the 6th textbook you could obtain at no cost containing examples from the idea of advanced features. during this quantity we will think about the principles of calculations or residues, either in finite singularities and in ∞. the idea 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 publication.

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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. 2 2 1 it e + e−it 2 + 1 1 · eit − e−it 2 2i Alternatively we may apply Rule III, thus res (f ; z0 ) = 6A B − 2AB 2 3 (B ) . If we put A(z) = ezt B(z) = z 2 + 1 and 2 = z 4 + 2z 2 + 1, then A(z) = ezt , A (z) = t ezt , B(z) = z 4 + 2z 2 + 1, B (z) = 4z 3 + 4z, B (z) = 12z 2 + 4, B (3) (z) = 24z, A(i) = eit , A (i) = t eit , B(i) = 0, B (i) = 0, B (i) = −8, B (3) (i) = 24i, A(−i) = e−it , B(−i) = 0, B (−i) = 0, B (−i) = −8, B (3) (−i) = −24i, hence, res ezt (z 2 + 1) 2 ;i = 6t eit · (−8) − 2eit · 24i 1 i = − t eit − eit , 3(−8)2 4 4 and res ezt (z 2 + 1) 2 ; −i = 6t e−it · (−8) − 2e−it · (−24i) 1 i = − t e−it + e−it , 2 3(−8) 4 4 and we proceed as above.

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 where ∗ z dz = − z4 − 1 ∗ |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. wanted: ambitious people Please click the advert At NNE Pharmaplan we need ambitious people to help us achieve the challenging goals which have been laid down for the company.

2 1 –2 –1 0 1 2 –1 –2 Figure 1: The path of integration C2 and the poles on the imaginary axis. (d) The path of integration Ca passes through the two removable singularities z0 = 0 and z1 = i. com 30 Complex Funktions Examples c-6 Residues in finite singularities is defined, and we get by Cauchy’s residue theorem that the value is given by Ca i tanh π z dz = 2π i res f (z) ; z(z − i) 2 = 2π i res (f (z) ; z0 ) = 2π i · 4 = 8i, π where we have used that z0 is the only pole inside Ca , and where res(f (z) ; zn ) has been computed in (b).

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