(Free42) roundoff for complex SQRT

[Dan]

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:





The second image shows the value of the real and imaginary parts of the square root in memory (2nd line from top). I know nothing about the 42S, but I know early HP calculators used the "Friden" algorithm for square root. Is it possible the 42S did too?