(Free42) roundoff for complex SQRT

[Claudio L.]

(04-04-2018 10:24 AM)Dan Wrote:  I used the "Friden" algorithm for square root on the AriCalculator, and sqrt(z) = sqrt(.5(a+sqrt(a^2+b^2)) )+ sqrt(0.5(-a+sqrt(a^2+b^2))). This gives exact results for 8i and other z I tried for which Re(z) = 0 and Im(z)/2 is a perfect square

This is what I used on newRPL too, but I'm probably going to change it to what Werner showed above for the 48, it saves one square root, so it's roughly 30% faster.