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Julia Set Z^-9+c(c=-0.9305i)

LEleszpio•Created July 30, 2019
Julia Set Z^-9+c(c=-0.9305i)
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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=-0.9305i

Project Details

Project ID322683879
CreatedJuly 30, 2019
Last ModifiedMarch 10, 2021
SharedMarch 10, 2021
Visibilityvisible
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