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

[ijabbott]

(10-25-2018 04:43 PM)sasa Wrote:  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.

So you're objecting to it factoring out the radical part as a real number? I think \(\frac{1+i}{\sqrt{2}}\) is easier to visualize than \(\sqrt{i}\).