sqrt question - Printable Version +- HP Forums ( https://www.hpmuseum.org/forum)+-- Forum: Not HP Calculators ( /forum-7.html)+--- Forum: Not remotely HP Calculators ( /forum-9.html)+--- Thread: sqrt question ( /thread-8126.html)Pages: 1 2 |

RE: sqrt question - Claudio L. - 04-26-2017 03:08 AM
(04-24-2017 09:36 PM)David Hayden Wrote:(04-09-2017 03:59 AM)Claudio L. Wrote: sqrt(1) = 1Isn't it: You started reading the thread from the bottom up :-) That post was intended as proof that even when selecting always the positive root as "the one true answer", it is possible to arrive to the other root by using trivial algebraic manipulations. Otherwise, I agree with you and Han that both answers are correct and valid, one is not better than the other. RE: sqrt question - Han - 04-28-2017 06:09 PM
(04-09-2017 03:59 AM)Claudio L. Wrote:(04-07-2017 10:54 PM)Han Wrote: Is this due to the sqrt() function, though? This seems like a consequence of assuming factorization properties of 1 and -1 that may not still hold true for complex numbers. It is not clear to me what you mean by the factorization holds true for complex numbers. I agree that \( \sqrt{ab} = \sqrt{a} \sqrt{b} \) provided that \( a \), \( b \), and \(ab \) are non-negative. However, I question whether the definition of \( \sqrt{x} \) has been implicitly changed when you allow \( a \) and \(b \) to be negative. For complex numbers, which can be represented as \( re^{i\theta} \), (where \( r \) is a non-negative real number and \(-\pi < \theta \le \pi \) ), we have the "principal root" \[ \sqrt{z} = \sqrt{re^{i\theta}} = \sqrt{r} e^{i\theta/2} \] The reason your example produces two outcomes is because you did not define the square root function (over the complex plane) to be one-to-one (unless you are restricting \(\theta \) to be strictly positive and less than or equal to \( 2\pi \) ). My point here is that it is mathematically possible to define the square root function for a complex number without obtaining ambiguous results. |