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Extra resources for Complex Functions Examples c-2 – Analytic Functions
Z = |z| ⎩ Limits 1 for y = 0 and x > 0, −1 for y = 0 and x < 0, we cannot extend this function continuously to z = 0. g. ⎧ ⎨ Re z 2 x2 − y 2 = = ⎩ x2 + y 2 |z|2 1 for y = 0 and x = 0, −1 for x = 0 and y = 0, we cannot extend this function continuously to z = 0. (d) Since Re(z) = |z| |x| x2 + y 2 ≤1 for every z = 0, we get z Re(z) z Re(z) −0 = ≤ |z| · 1 → 0 |z| |z| for z → 0, and we conclude that this function can be extended continuously to z = 0 with the value f (0) = 0. 3 Check if the following limit values exist.
A) By using the Theory of Complex Functions we get 4z 3 dz = z 4 C 3 3 = 0. Alternatively it follows by using the parametric description, 1 4z 3 dz = 0 C 4 · 33 · e6iπt · 3 · 2iπ · e2iπt dt = 34 1 0 (b) We get by insertion of the parametric description that 1 z dz = 0 C 3 e−2iπt · 3 · 2iπ · e2iπt dt = 18πi. (c) We get by insertion of the parametric description that C 1 dz = z 1 0 1 · 3 · 2iπ e2iπt dt = 2iπ. 3 e2iπt 56 8iπ · e8iπt dt = 81 · e8iπt 1 0 = 0. 10 Sketch the curve C of the parametric description z = e4iπt , t ∈ [0, 1], and indicate its orientation.
We shall everywhere use the parametric description z = eit , t ∈ [0, 2π], 47 Complex Funktions Examples c-2 Line integrals of the curve C. Then dz = i eit dt, and we get: (a) C dz = z C dz = |z| C dz = z2 C dz = |z 2 | 2π 2π i eit dt = i eit 0 dt = 2πi, 0 (b) 2π i eit dt = |eit | 0 2π i eit dt = eit 0 2π 0 = 1 − 1 = 0, (c) 2π i eit dt = e2it 0 2π i e−it dt = −e−it 0 2π 0 = −1 − (−1) = 0, (d) 2π 0 i eit dt = |e2it | 2π i eit dt = eit 0 2π 0 = 1 − 1 = 0. 3 Find the value of the complex line integral C |z| dz, when the curve C is (a) the line segment from −i to i, (b) the left half of the unit circle run through from −i to i, (c) the right half of the unit circle run through from −i to i.