ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Thu, 05 Dec 2019 23:50:40 +0100Equation in complex numbershttps://ask.sagemath.org/question/48956/equation-in-complex-numbers/ I need to solve the following equation.
solve(z^2 == (1-sqrt(3)*I)*z.conjugate(), z)
Sage says
[z == -sqrt((-I*sqrt(3) + 1)*conjugate(z)), z == sqrt((-I*sqrt(3) + 1)*conjugate(z))]
I'd like to get solutions in polar form, something like
[z == 0, z == 2*e^(pi*I/9), z == 2*e^(7*pi*I/9), z == 2*e^(13*pi*I/9)]
Or I'd like to get the absolute values and arguments of the solutions. Is it possible in Sage?EvgenyMThu, 05 Dec 2019 23:50:40 +0100https://ask.sagemath.org/question/48956/