Question

y^2=-i; where i is a imaginary root. y?

Answer 1

if you imagine y as \[r*(\cos(\theta)+i \sin (\theta))\] then :
r=+1 or -1
\[\theta\]=\[-\pi/4\]

Answer 2

Let \[y=a+ib\] then \[y^2=a^2+2aib-b^2=-i\] hence \[a^2-b^2=0, 2ab=-1\] hence \[b=-1/(2a)\] \[a^4=1/4\] \[a=\pm \sqrt{1/2}\] then \[y=(1/\sqrt{2})(1-i)\] or \[y=(1/\sqrt{2})(-1+i)\]

Answer 3

got it?