By Mejlbro L.
This is often the second one textbook you could obtain at no cost containing examples from the speculation of complicated features. there'll even be examples of complicated capabilities, advanced limits and complicated line integrals.
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Extra info for Complex Functions Examples c-2 - Analytic Functions
Along the quarter circle we use the parametric description Γ : ζ(t) = 1 + A · eit , t ∈ 0, π , 2 hence we get the estimate eiz Γ z+ 1 z ≤ dz ≤ π 2 π 2 exp i 1 + A eit A2 · A dt ≤ 2 exp(−A sin t) dt 1 A −1 0 0 A− A π 2 2 π 1 2 A2 A · · 1 − e−A → 0 for A → +∞, exp − A · t dt ≤ 2 2 A −1 0 A −1 2 A π where we have used that 2 ≤ sin t ≤ t π for t ∈ 0 , π , 2 which follows easily by considering the graph of sin t in the given interval. This proves that − sin t ≤ − 2 t π for t ∈ 0 , π , 2 which was used in the estimates above.
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 1 0 = 1 − 1 = 0. (b) By insertion of the parametric description we get 1 z dz = C 0 e+iπt · (−iπ)e−πt dt = −iπ. (c) By insertion of the parametric description we get C 1 dz = z 1 0 1 (−iπ)e−iπt dt = −iπ.
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. com 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.