Free42 and cube root of complex number

[Albert Chan]

(02-24-2021 02:02 AM)Thomas Okken Wrote:  atan2(-0, -8) => -pi, while atan2(0, -8) => pi

-pi < arg(z) ≤ pi

Since polar form and rectangular form should match, I think -8 ± 0i should be adjusted to -8 + 0i = 8 exp(i*pi)

For the same reason, sqrt(-4 - 0i) = sqrt(-4 + 0i) = 2i

4 [ENTER] 0 [COMPLEX] [+/-] [SQRT]       → 2i