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Julia Set Z^9+c(c=-0.879+0.235i)

LEleszpio•Created July 9, 2019
Julia Set Z^9+c(c=-0.879+0.235i)
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Z=x+iy Z=|Z|(cosα + i sinα) where |Z|=sqrt(x^2+y^2) α=arctg(y/x) To calculate the power of complex numbers, I used the de Moivre formula. Z^9+c=(|Z|^9)*(cos(9*α) + i sin(9*α))+c c=-0.879+0.235i

Project Details

Project ID320303890
CreatedJuly 9, 2019
Last ModifiedJanuary 18, 2020
SharedJanuary 18, 2020
Visibilityvisible
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