HP50g simplifing a root

[Albert Chan]

(10-11-2020 06:28 PM)Albert Chan Wrote:  Example, simplify ³√(36 + 20i√7)

>>> from mpmath import *
>>> A, B, k = 36, 20, -7
...
>>> c = cbrt(A*A-B*B*k); print c
16.0

Here is a novel way to solve for r instead, then get a.
(Note: b can be halves, just like a)

R/r = (B/b)² = (3c + 4r)² = integer, where r = b²k

For this case, we hit the jackpot. If b=±1, 3c+4r = 3c+4k = B

r = k = -7
a = A / (c+4r) = 36/ (16-28) = -3

3*arg(a±√r) = 3*atan2(±√7, -3) ≈ ±2.3098 pi

We need to match above angle to arg(A+√R), which is on the first quadrant.
+2.3098 pi - 2 pi = 0.3098 pi matches.

³√(36 + 20i√7) = (-3 + i√7) / (-1)^(2/3)

This is just for fun. No need to solve the cubics.
Directly calculate ³√(A+√R) (previous post) is preferred.

Comment: Solving for b instead, we can take square root of both side

B/b = 3c + 4r = 3c + 4kb² = integer

This removed the ambiguity of b = ±1 to b = B/(3c + 4kb²) = 1
Above example need to match angles anyway, so I skipped this.