Julia Set Z^11+c(c=-0.866+0.123i)

LEleszpio•Created July 13, 2019

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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^11+c=(|Z|^11)*(cos(11*α) + i sin(11*α))+c c=-0.866+0.123i

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Project ID320817496

CreatedJuly 13, 2019

Last ModifiedSeptember 18, 2019

SharedSeptember 18, 2019

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