HP50g simplifing a root

10122020, 03:17 AM
(This post was last modified: 10132020 12:08 PM by Albert Chan.)
Post: #25




RE: HP50g simplifing a root
(10112020 06:28 PM)Albert Chan Wrote: Example, simplify ³√(36 + 20i√7) 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/ (1628) = 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. 

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