# Solving an equation with complex parameters in Wolfram|Alpha

Posted 3 months ago
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 If one has an equation of the form: (y + i /2) g(u, w) = 0, where i denotes √(-1), where y is real while u & v are complex, and where g(v, w) is an algebraic function of v and w; then: is the correct solution for y of this equation: 'no solution for y' {as is given by Wolfram Alpha Pro for a special case of the algebraic function denoted by g(u, w)} ... or else is the correct solution: 'y = any real number'? This strange problem arose during my recent correspondence with a Prof. of Mathematics at my alma mater of UC Berkeley, who specializes in algebra & number theory & who argued that one could divide both sides of this equation by (y + i / 2) since y is defined to be real and so (y + i / 2) cannot vanish; and then he argued further that therefore one obtains (after this division) the algebraic equation: g(v, w) = 0 which is, of course, trivially satisfied for all real y since y does not appear in it, and so this equation places no constraints on (assumed real) y at all ... whatsoever. However, this seems to me to be very poor methodology, does it to you as well? (And please give me a good supporting reference or two to your answer to this question.)
 Maybe "no solution for y" means that there is no formula needed for y because y is arbitrary, not that there is no value for y that satisfies the equation. For example, the equation (1 + y^2) x == 0 among the reals is equivalent to x == 0, which does not contain y: in the cartesian plane the set of solutions is the whole y axis.