Proving a complex number has exactly two solutions

Questions
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1

Prove algebraically that for any complex number w?0 the equation z^2=w has exactly two solutions.

When I sub in x+iy for w, I get z^2 = x + iy. Knowing z is complex, then I sub in x+iy for that as well to get x^2+2xiy+(iy)^2 = x + iy. Subtracting over I get x^2-x+2xiy+(iy)^2-iy. From here I am not sure what to do or if I am doing it right. Any help would be greatly appreciated! Thanks!

2

One must not substitute z → x + iy as well as w → x + iy, that paves the way into confusion. Substitute instead z → x + iy, w → u + iv and go ahead, express u and v by the real and imaginary part of the left hand side.