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Julia Set Z^-3+c(c=-0.7887)

LEleszpio•Created August 6, 2019
Julia Set Z^-3+c(c=-0.7887)
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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^-3+c=(|Z|^-3)*(cos(-3*α) + i sin(-3*α))+c c=-0.7887i

Project Details

Project ID323361931
CreatedAugust 6, 2019
Last ModifiedAugust 8, 2019
SharedAugust 6, 2019
Visibilityvisible
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