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Then |z|=1 sin z sin z 2π cosh 1 · 2π ≤ dz ≤ max < 2πe. 1 We shall here demonstrate – although it will later follow more systematically – that |z|=1 sin z dz = 2πi. z2 First method. If one already knows a little of calculus of residues, then the task is quite simple: We sin z see that z = 0 is a simple pole of , because sin z has a simple zero for z = 0, and because z2 z = 0 is the only singularity inside the circle |z| = 1. Hence by the theorem of residues, |z|=1 sin z ;0 z2 sin z dz = 2πi res z2 = 2πi lim z→0 sin z = 2πi.

Com 31 Complex Functions Examples c-3 The exponential function and the logarithm function Hence, the streamlines are given by ψ(x, y) = ex sin y = c > 0, thus c sin y x = ln , y ∈ ]0, π[, c ∈ R+ , because it is here easier to express x as a function of y than vice versa. 3 Prove that √ 2 2 exp πi 12 = √ 3+1 +i √ 3−1 . Hint: Apply 1 1 1 = − . 12 3 4 Using the hint we get √ 2 2 exp πi 12 √ = 2 2 · exp √ = 2 2· √ πi πi π π π π · exp − = 2 2 · cos + i sin · cos − i sin 3 4 3 3 4 4 √ √ √ √ 1 1 3 1 √ −i√ = 1 + i 3 (1 − i) = 1 + 3 + i 3−1 .

So summing up the solution can be written z= π + 2pπ ± 2 √ π − i ln( 2 + 1) , 2 p ∈ Z. Please click the advert what‘s missing in this equation? You could be one of our future talents MAERSK INTERNATIONAL TECHNOLOGY & SCIENCE PROGRAMME Are you about to graduate as an engineer or geoscientist? Or have you already graduated? P. Moller - Maersk. 14 Find all solutions of the equation tan z = 3i, in the form x + iy. The solutions are given by z = i − 3i i + 3i 1 −2i log 2i 4i 1 π i {− ln 2 + i(π + 2pπ)} = + pπ + ln 2, 2i 2 2 = arctan(3i) = 1 log 2i = = 1 1 log − 2i 2 p ∈ Z.

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Complex Functions Examples c-3 - Elementary Analytic Functions and Harmonic Functions by Leif Mejlbro


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