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Julia Set Z^-8+c(c=-0.899)

LEleszpio•Created July 22, 2019
Julia Set Z^-8+c(c=-0.899)
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Description

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^-8+c=(|Z|^-8)*(cos(-8*α) + i sin(-8*α))+c c=-0.899

Project Details

Project ID321876618
CreatedJuly 22, 2019
Last ModifiedJuly 24, 2019
SharedJuly 22, 2019
Visibilityvisible
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