complex numbers - How to solve the equation $z^2 = i-1$?

$\ z^2 = i-1 \ $
Hey guys, I couldn't find a way to solve this problem. The question suggests I replace $\ z\ $ with $\ x+iy\ $ and then go from there, but I always end up having to complete the square and end up with a completely different answer to the back o the book. Can you please help?
Thanks

Answer

let

$$z=x+iy$$ then we get
$$x^2-y^2+2xyi=i-1$$ so we get
$$x^2-y^2+1=0$$
and $$2xy-1=0$$
Can you solve this?