Given:
z^2-i=0
z=x+iy
z^2=(x+iy)^2
z^2=x^2+2ixy+(iy)^2
i^2=-1
z^2=(x^2-y^2)+2ixy
substituting for z^2
(x^2-y^2)+2ixy-i=0
(x^2-y^2)+(2xy-1)i=0+0i
Equating the real and imaginary parts
x^2-y^2=0
2xy-1=0
2xy=1
y=1/(2x)
x^2-(1/(2x))^2=0
x^2-1/(4x^2)=0
Let
t=x^2
t-1/(4t)=0
4t^2-1=0
4t^2=1
t^2=1/4
t=+-1/2
t=1/2
t=-1/2
ie
x^2=1/2
x=+-1/sqrt2
x=1/sqrt2
x=-1/sqrt2
x^2=-1/2
x=+-i/sqrt2
x=i/sqrt2
x=-i/sqrt2
Thus,
Substituting for x,
x=1/sqrt2 ,
y=1/(2x)=1/(2xx1/sqrt2)
y=1/sqrt2
x=-1/sqrt2 ,
y=1/(2x)=1/(2xx-1/sqrt2)
y=-1/sqrt2
x=i/sqrt2 ,
y=1/(2x)=1/(2xxi/sqrt2)
y=-i/sqrt2
x=-i/sqrt2 ,
y=1/(2x)=1/(2xx-i/sqrt2)
y=i/sqrt2
We have
(1/sqrt2,1/sqrt2)
(-1/sqrt2,-1/sqrt2)
(i/sqrt2,-i/sqrt2)
(-i/sqrt2,i/sqrt2)
as solutions
z=1/sqrt2+1/sqrt2i
z=-1/sqrt2+-1/sqrt2i
z=i/sqrt2+1/sqrt2
z=-i/sqrt2-1/sqrt2
Rearranging
z=1/sqrt2+i/sqrt2
z=-1/sqrt2-i/sqrt2
z=1/sqrt2+i/sqrt2
z=-1/sqrt2-i/sqrt2