Free42 and cube root of complex number

[Albert Chan]

(02-24-2021 01:35 PM)Thomas Okken Wrote:  

(02-24-2021 08:59 AM)Werner Wrote:  Please, leave it as it is - Free42 would be one of the only ones to get it right ;-)

There is that! But the HP-42S doesn't support -0 ...

Hi, Thomas Okken

Can you elaborate what is decided ?

If Free42 do support -0, then sqrt(-4-0i) = sqrt(4*exp(-pi*i)) = 2*exp(-pi/2*i) = -2i

But, user need to know "-0" is actually inputed.
-0 should displayed as "-0", not "0"
-4-0*i should displayed as "-4 -i0", not "-4 i0"

If not, (-8-0i)^(1/3) = (-8+0i)^(1/3) = 1 + sqrt(3)*i