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By Leif Mejlbro

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Com 57 Complex Functions Examples c-3 Trigonometric and hyperbolic functions Finally, we get the solution z= ⎧ 1 √ √ ⎪ ⎪ ⎨ i log( 2 − 1) = 2pπ − i ln( 2 − 1), p ∈ Z, ⎪ ⎪ ⎩ 1 log(−√2 + 1) = π + 2pπ − i ln(√2 + 1), i √ √ It follows from ( 2 − 1)( 2 + 1) = 1 that √ √ ln( 2 − 1) = − ln( 2 + 1), p ∈ Z. 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?

Thus z 3 = (x + i y)3 = x3 + 3i x2 y − 3xy 2 − i y 3 = x3 − 3xy 2 + i 3x2 y − y 3 , and z 2 = (x + i y)2 = x2 − y 2 + i · 2xy. It follows immediately that 3x2 y − y 3 = Re −i z 3 and i xy = Re − z 2 , 2 so u(x, y) = Re −i z 3 − i 2 z , 2 and we conclude that v(x, y) = Im −i z 3 − i 2 z 2 + c = −Re z 2 + 1 2 z 2 + c = −x3 + 3xy 2 − 1 2 1 2 x + y + c, 2 2 and f (z) = −i z 3 − i 2 z + i c, 2 c ∈ R. com 64 Complex Functions Examples c-3 Harmonic functions 2) Alternatively we conclude from Cauchy-Riemann’s equations that v(x, y) ∂u ∂u dx + dy ∂y ∂x − 3x2 − 3y 2 + x dx + (6xy + y)dy = c+ − = c+ −3x2 dx + 3y 2 dx − x dx + 6xy dy + y dy = c+ d −x3 + 3y 2 dx+3x d y 2 = c+ d −x3 + 3y 2 x − = −x3 + 3y 2 x − =c+ 1 1 − d x2 + d y 2 2 2 1 2 1 2 x + y 2 2 1 2 1 2 x + y + c, 2 2 c ∈ R, and f (z) = u + i v = 3x2 y − y 3 + xy + i −x3 + 3y 2 x − = −i x3 +3x2 (i y)+3x(i y)2 +(i y)3 − i 2 z + i c, 2 x2 −y 2 +2i xy +i c c ∈ R.

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