(Free42) roundoff for complex SQRT

[Werner]

Nitpicking, perhaps, but if the original 42S gets it right - then so should Free42 ;-)
Try SQRT(a+ib) with a=0 and b/2 a perfect square eg. b=8
Free42 shows 2+i2, which SHOW reveals to be really 2+i(2-1E-33).
Almost any b you throw at it has a rounding error like that, in the real or complex part.
The 12-digit 42S gets the exact result - it must have special code for a=0, and return SQRT(ABS(b)/2)*(1+i*SIGN(b)) in that case - even when b/2 would underflow.

Cheers, Werner