Multiple (possibly complex) roots?

[unfrostedpoptart]

Hi all!

Thanks to Jonathan Cameron, I've got my working WP34S (with crystal!) now. It's amazing and I was going through the manuals (starting with the beginner's guide) since it's been many years since I did a lot with an HP calculator. Some great surprises, like the fraction modes!

Question, though, about complex math and/or roots in general. The nth root of a number has n solutions. If you plot them on a complex graph, they'll all fall in a circle and be equidistant (same angles). Simplest is square root of 1. Answers are 1 and -1. SRT -1 = i and -i. Cube root of 1 = 1, -0.866 + 0.5i, and -0.866 - 0.5i.

On the HP/WP, which root does it normally pick? I assume to get all of them, I'd have to use SLV with different starting points. I guess as wonderful as the calc is, it's not Matlab or Mathematica

[Thomas Klemm]

(04-21-2014 01:20 AM)unfrostedpoptart Wrote:  On the HP/WP, which root does it normally pick?

It usualy picks the root with the smallest argument. Thus the WP-34S returns -2 for \(\sqrt[3]{ -8}\) but returns \(1+i\sqrt{3}\) when using [CPX] \(\sqrt[3]{-8+i0}\).

Quote:I assume to get all of them, I'd have to use SLV with different starting points.

When using Newton's method the choice of the starting point is crucial. The basin of attraction (the set of initial points that converge to the same root) for each root is a fractal:

Make sure to start in the vicinity of one of them.

I'm afraid that SLV can't be used to find complex roots. For polynomials with real coefficients you could use Bairstow's Method.

Cheers
Thomas

[Thomas Klemm]

This is a nice animation for the polynomial \(x^5 -x-\frac{t}{100}\) where t is the frame count.