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?