z^n = a + bi
Author: Yoichiro1224Executions: 148
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This formula solves the equation "z^N = a + b*i."
z = (R^(1/N)) * exp(i * (2πk + α) / N)
where k = 0, 1, 2, ..., N-1, R = √(a² + b²), and α = atan(b/a)
where k = 0, 1, 2, ..., N-1, R = √(a² + b²), and α = atan(b/a)
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