Solving sqrt(i)=z, one or two solutions?

[sasa]

(10-25-2018 10:52 AM)parisse Wrote:  I wonder how can sasa pretend that the Prime answer is unexpected.

Actually, I do not... Here what is a deal:

if we have: \( \sqrt{i}=z \), it is the first degree polynomial and actually exact solution, as sqrt from any constant is again a constant - there is nothing to simplify. Hence, we fully agree that the (i) is a constant. This way it is pointless to do any solving, as well as fundamental theorem state. Actually, if we need to "transform" \( \sqrt{i} \), exact alternative would be: (\(-1)^{\frac{1}{4}}\). Alternate form as \(\frac{1+i}{\sqrt{2}}\) is not quite correct - anyone is free to prove differently.

if we have: \( i=z^{2} \) we know that it is second degree polynomial with two solutions.

Simply just another sample from the twilight zone...