Complex Functions Examples c-2 - Analytic Functions by Leif Mejlbro

By Leif Mejlbro

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Then compute 4z 3 dz, (a) z dz, (b) C (c) C C 1 dz. z 3 2 1 –3 –2 –1 0 1 2 3 –1 –2 –3 Figure 35: The curve C and its orientation. (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 8iπ · e8iπt dt = 81 · e8iπt 1 0 = 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.

Then we have the following computation, 1 4z 3 dz = 0 C = 4 1+ t2 + it 1 4 0 1 = 4 0 = 4 0 1 1+t2 3 3 (i + 2t)dt +3it 1+t2 2 −3t2 1+t2 −it3 (2t + i)dt t6 +3t4 +3t2 +1−3t4 −3t2 +i 3t5 +6t3 +3t−t3 2t7 +2t−3t5 −5t3 −3t +i t6 +1+6t6 +10t4 +6t2 1 8 1 6 5 4 1 2 t − t − t − t + i t7 +2t5 +2t3 +t 4 2 4 2 = 1 − 2 − 5 − 2 + 4i(1 + 2 + 2 + 1) = −8 + 24i. (2t+i)dt dt 1 = 4 0 (b) When we insert the parametric description we get 1 z dz = 0 C = 1 + t2 − it (2t + i) dt = 1 4 3 2 1 t + t + i − t3 + t 2 2 3 1 0 1 2t3 + 2t + t + i t2 + 1 − 2t2 =2+ 0 dt 2 i.

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