By Leif Mejlbro

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**Additional info for Complex Functions Examples c-2 – Analytic Functions **

**Sample text**

A) The sequence zn = in converges towards the limit value 0. In fact, +i n2 0 ≤ |zn − 0| = |zn | = √ n n4 n 1 n ≤ √ = 2 = →0 4 n n +1 n for n → +∞. (b) Since this sequence repeats cyclically the numbers 1+i √ , 2 i, −1 + i √ , 2 −1, −1 − i √ , 2 −i, 1−i √ , 2 1, it is divergent. (c) The sequence zn = 1+i 4 0 ≤ |zn − 0| = |zn | = n √ 2 4 converges towards the limit value 0. In fact, n →0 for n → +∞. 42 Complex Funktions Examples c-2 Limits (d) Since n2 + n = n(n + 1) in an even number for every n ∈ N, we see that zn = exp i n2 + n π = exp(2ipπ) = 1 for every n ∈ N, where 2p = n(n + 1).

2 (c) Using the Theory of Complex Functions we get C π 1 1 dz = [Log z]1+i = Log(1 + i) − Log 1 = Log(1 + i) = ln 2 + i . 1 4 2 z Alternatively we get by the parametric description that C 1 dz z 1 1 i(1 − it) 1 dt = i dt + 2 1 + t 1 + t2 0 0 0 π 1 1 1 ln 1 + t2 0 = ln 2 + i . 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.

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. This is in particular true for p = 0. 8 Sketch the curves u(x, y) = constant and v(x, y) = constant in the z-plane for the following functions: (a) f (z) = 1 , z (c) f (z) = (1 − 2i)z.