Complex Functions Examples c-2 – Analytic Functions - download pdf or read online

By Leif Mejlbro

ISBN-10: 8776813843

ISBN-13: 9788776813840

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Complex Funktions Examples c-2 Line integrals We stop the computations at this point because they should only serve as an illustration of the fact that in the Theory of Complex Functions it is worth always to look for alternatives which might be easier to apply. It is of course possible to reduce the computations above to the result 1 1 ln 5 + i Arctan , 2 2 by starting the reduction by 4 √ = 6+2 5 2 √ = 3+ 5 √ 2 2 5+1 =√ 2 , 5+1 and analogously. 12 Compute C (z + 1)dz, where C is that part of the parabola of the equation y = x2 , which starts at z = 0 and ends at z = 1 + i.

5 Compute the line integral z = 1 to z = i. C iπ . 2 x2 + i y 3 dz, where C is the straight line segment from A parametric description of C is given by x(t) = 1 − t, y(t) = t, t ∈ [0, 1]. 5 Figure 30: the path of integration C. 5 Figure 31: The path of integration C with its direction. Hence, by insertion x2 + i y 3 dz 1 = 0 C 1 = 0 = 0 = = 1 x(t)2 + i y(t =3 · {x (t) + i y (t)} dt (1 − t)2 + it3 · (−1 + i) dt −(1 − t)2 − t3 dt + i 1 1 0 (1 − t)2 − t3 dt 1 t4 t4 (t − 1)3 (t − 1)3 − − +i − 3 4 0 3 4 0 1 7 1 1 1 1 i.

32 Complex Funktions Examples c-2 Complex Functions Now, x = eu + u runs through all of R, when u runs through R, so the curve is the horizontal line y = 2pπ. This is in particular true p = 0, so in this case the curve is the whole of the x-axis. If v = (2p + 1)π, p ∈ Z, then x = u − eu and y = (2p + 1)π, u ∈ R. Since dx = 1 − eu , du we conclude that the function x(u) has a maximum for u = 0, corresponding to x = −1, and since x(u) → −∞ for u → +∞ and for u → −∞, we conclude that the half lines x ≤ −1, y = (2p + 1)π, p ∈ Z, are run through twice.

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Complex Functions Examples c-2 – Analytic Functions by Leif Mejlbro


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