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By Leif Mejlbro

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By separation into real and imaginary part we get x = u + eu cos v, y = v + eu sin v. 6 4 2 –1 0 2 1 3 –2 –4 –6 Figure 18: The images of the curves u = k for k = −1, 0 and 1. If u = k, and v ∈ R is considered as a parameter, we get the parametric description x = k + ek cos v, y = v + ek sin v, v ∈ R, 31 Complex Funktions Examples c-2 Complex Functions 6 5 4 3 2 1 0 –2 2 6 4 8 Figure 19: The images of the curves v = k for k = 0, π π 3π and π. , , 4 2 4 which cannot be further reduced. If v = k, and u ∈ R is considered as a parameter, then x = u + eu cos k, y = k + eu sin k, u ∈ R.

9 Sketch the curves u(x, y) = constant and v(x, y) = constant in the z-plane for the following functions, (a) f (z) = z 2 , (b) f (z) = z + z 2 , (c) f (z) = z+i . z−i 2 1 –2 0 –1 1 2 –1 –2 Figure 24: The level curves of (a). (a) We get by a separation into real and imaginary part, u(x, y) = x2 − y 2 og v(x, y) = 2xy. The level curves u = k form a family of hyperbolas and the straight lines y = x and y = −x. The level curves v = k are also a family of hyperbolas with the axes added. We see that apart from in the singular point (0, 0), every curve from one system of curves is always orthogonal to any curve from the other system of curves.

8 Sketch the curve C of the parametric description z = e−iπt , t ∈ [0, 1], and indicate its orientation. Then compute 4z 3 dz, (a) z dz, (b) C (c) C C 1 dz. 2 Figure 34: The curve C with its orientation. (a) By using the Theory of Complex Functions we get 4z 3 dz = z 4 C −1 1 = (−1)4 − 14 = 0. Alternatively we apply the parametric description 1 4z 3 dz = 0 C 4 e−3iπt · (−iπ)e−iπt dt = 1 (−4iπ)e−4iπt dt = e−4iπt 0 (b) By insertion of the parametric description we get 1 z dz = 0 C e+iπt · (−iπ)e−πt dt = −iπ.