New PDF release: Complex Functions Examples c-2 – Analytic Functions

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Sin k forudsat, at Hence we obtain the explicit expression of the curve, x = ln y−k sin k + y−k · cos k = ln sin k y−k sin k + (y − k) cot k for y−k > 0. sin k 6 4 2 –3 –2 –1 1 2 3 –2 –4 –6 Figure 20: The images of the curves v = k for k = −2π, −π, 0, π and 2π. If v = 2pπ, p ∈ Z, then x = u + eu and y = 2pπ, u ∈ R. 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.

E. a field around a dipole at the point (0, 1). We get for the level curve u(x, y) = 1 + k, k = 0, that k= 2(y − 1) , x2 + (y − 1)2 (x, y) = (0, 1), thus x2 + (y − 1)2 − 1 2 1 (y − 1) + 2 = 2 , k k k (x, y) = (0, 1), which we also write x2 + y − 1 − 1 k 2 = 1 , k2 (x, y) = (0, 1). This is the equation of a circle of centrum 0, k+1 k and radius 1 , with the exception of the |k| singular point (0, 1). The case v(x, y) = k is treated analogously. e. the y-axis), with the exception of the singular point (0, 1).

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. (b) We first compute u + iv = z + z 2 = x2 − y 2 + x + i(2xy + y). We get by a separation into real and imaginary part, u(x, y) = x2 + x − y 2 = x+ 1 2 2 − y2 − 1 4 36 Complex Funktions Examples c-2 Complex Functions 2 1 –3 –2 –1 0 1 2 –1 –2 Figure 25: The level curves of (b). and v = 2xy + y = 2y x + 1 2 , thus the curves are the same as in (a), only the centrum has been translated to 37 1 − ,0 .

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