(Free42) roundoff for complex SQRT

[Thomas Okken]

(02-12-2021 02:45 PM)Werner Wrote:  When X = 0+iB, SQRT(X) should have equal real and imaginary parts (apart from the sign).
But that is (still) not the case.

I don't think that is something I ever tested. I implemented your algorithm in 2.0.21 and verified that it gave the desired result for 0+iB when B/2 is a perfect square... I suppose this means an explicit check for Re(X)=0 will be needed after all.