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11 Find all complex solutions of the equation {(1 + i) cos z + (1 − i) sin z}4 = 16i. First we get (1 + i) cos z + (1 − i) sin z = = = √ π 2 π · eiz − e−iz exp i · eiz + e−iz − i exp −i 4 2 4 1 π π √ exp i · eiz + e−iz − exp i · eiz − e−iz 4 4 2 √ π 1 π √ exp i −z . · 2 e−iz = 2 · exp i 4 4 2 Then by insertion into the equation, {(1 + i) cos z + (1 − i) sin z}4 = 4 exp(i{π − 4z}) = −4 e−4iz = 16i, and the equation is reduced to e4iz = − 4 i π = = exp −2 ln 2 + i + 2pπ 16i 4 2 p ∈ Z. , Then by taking the logarithm, 4iz = −2 ln 2 + i π + 2pπ , 2 p ∈ Z, and the complete solution becomes zp = π π i + p + ln 2, 8 2 2 p ∈ Z.

The function f (z) = Log z 2 + 2 is analytic in A, so it follows by the chain rule that f (z) = 2z , +2 z2 z ∈ A. 2) Let γ : [0, 1] → A be a parametric description of any differentiable curve from 0 ∈ A to −1 + i ∈ A. The integrand is equal to f (z), found above, so the primitive is given by f (z) dz γ = f (γ(1)) − f (γ(0)) = f (−1 + i) − f (0) = Log (−1 + i)2 + 2 − Log 02 + 2 = Log(−2i + 2) − Log 2 = Log(1 − i) = 1 π ln 2 − i , 2 4 Please click the advert where we have used that Log 2 = ln 2. 7 Define F (z) = exp z 2 , z ∈ C.

If y ∈ ]0, π[, then ψ(x, y) > 0, and the streamlines are given by ψ(x, y) = cosh x · sin y = c ∈ R+ . When we continue our investigation, we must split into the three cases, x ∈ ]0, 1[, c = 1, and c ∈ ]1, +∞[. com 51 Complex Functions Examples c-3 Trigonometric and hyperbolic functions 4 3 y 2 1 –3 –2 0 –1 2 1 3 x –1 Figure 12: Sketch of the streamlines. 1) If c ∈ ]0, 1[, then y is expressed as functions of x by y = Arcsin c cosh x c . cosh x y = π − Arcsin and 2) If c = 1, then we get the two so-called separatrices, most easily described by x = ± Arcosh 1 sin y 1+ = ± ln 1 − sin2 sin y = ± ln 1 ± cos y sin y = ± ln cot y , 2 because ln 1 − cos y sin y = − ln 1 + cos y sin y , and we can combine the two ± signs into one.

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


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