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Proving a complex number has exactly two solutions

Posted 11 years ago
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