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Alternatively, f (z) = ∂u ∂v ∂u ∂u +i = −i = −6xy − i 3y 2 − 3x2 = 3u x2 − y 2 + 2i xy = 3i z 2 , ∂x ∂x ∂x ∂y and then by finding a primitive, f (z) = i z 3 + CD1 , C1 ∈ C. We conclude from Re(f (z)) = Re i z 3 + Re (C1 ) = u(x, y) + 0, that C1 is purely imaginary, so f (z) = i z 3 + i C, C ∈ R. 3 Check if the following functions u(x, y) are harmonic: (a) u(x, y) = x2 − y 2 + y, (b) u(x, y) = x3 − y 3 , (c) u(x, y) = 3x2 y − y 3 + xy, (d) u(x, y) = x4 − 6x2 y 2 + y 4 + x3 y − xy 3 . If u is harmonic, then find the corresponding harmonic conjugated functions v, and f (z) = u + i v.

It is almost obvious that u(x, y) + i v(x, y) = x3 − 3xy 2 + i 3x2 y − y 3 = z 3 = f (z). 2 Find the constant a ∈ R, such that the function u(x, y) = y 3 + ax2 y becomes harmonic. Then find an harmonic konjugated function v of u, and describe f (z) = u + i v as a function of z alone. Clearly, u ∈ C ∞ R2 for every a ∈ R, so u is harmonic, if and only if Δu = 0. com 62 Complex Functions Examples c-3 Harmonic functions so Δu is identical zero, if and only if a = −3, in which case u(x, y) = y 3 − 3x2 y.

Hence, it suffices to prove that F (z) = Log z for z ∈ D . This follows from F (z) = Log z + z · 1 − 1 = Log z. z 2) Clearly, Log z is analytic in D , and since F (z) is a primitive of Log z, we get Log z dz γ = [F (z)]i1 = 1 · Log 1 − 1 − i · Log i + i = 0 − 1 − i i · = −1 + π 2 +i π + i. 1 Find the real and imaginary parts of (a) cosh(πi), (b) ei , (c) sin(πi), (d) sinh(1 + i). (a) Since cosh(πi) = cos π = −1, we get Re(cosh(πi)) = −1 and Im(cosh(πi)) = 0. (b) Since ei = cos 1 + i sin 1, we get Re ei = cos 1 and Im ei = sin 1.

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