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Proving a complex number has exactly two solutions

Posted 10 years ago

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!

POSTED BY: John Long

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.

POSTED BY: Udo Krause

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