No that doesnt work, because the fourth root of i has to equal i squared when squared. I got to my answer by using basic math.Jimmy T. Malice said:I think the fourth root of -1 is still i.
(ix+y)^2=y^2-x^2+2ixy therefore
x-y=0
xy=1/2=x^2
x=(+/-)root(1/2)=1/(+/-)root(2)
so (i/(+/-)root(2)+1/(+/-)root(2))^2=i
root(i)=(+/-)(1+i)/(+/-)root(2)