# Solve gives solution with ComplexInfinity -> number

I am trying to solve a cumbersome equation using

Simplify[Solve[(d*Cos[Φ]/Sqrt[x (x + J2) (x + J3) (x + J4)])*(1/
x + 1/(x + J2) + 1/(x + J3) +
1/(x + J4) + (x + J2) (x + J3) (x + J4) +
x*J2 (x + J3) (x + J4) + x*J3 (x + J2) (x + J4) +
x*J4 (x + J2) (x + J3)) + x*k0 + k1 + x*k1 + k2 == 0]]


When I plug this into Mathematica 9 I get

{{k2 -> -k0 x - k1 (1 + x) +
d Sqrt[x (J2 + x) (J3 + x) (J4 + x)] (-3 - 1/(
x^2 (J2 + x) (J3 + x) (J4 + x)) +
x (1/(J2 + x) + 1/(J3 + x) + 1/(J4 + x)) - (
1 + 1/((J2 + x) (J3 + x) (J4 + x)^2) +
1/((J2 + x) (J3 + x)^2 (J4 + x)) +
1/((J2 + x)^2 (J3 + x) (J4 + x)))/
x) Cos[Φ]}, {ComplexInfinity ->
x (J2 + x) (J3 + x) (J4 + x)}}


Does anyone know what it means for "ComplexInfinity-> x"? How can complex infinity approach a number? Is this some sort of asymptotic statement?

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It is a good idea to read the code formatting section of the help centre. –  Sektor Feb 27 '14 at 20:01
It appears to be a bug. That said, you should use the documented syntax for solve and specify the variable you want to solve for. See the documentation for details. –  Szabolcs Feb 27 '14 at 20:14
Solve needs two arguments, and you have only given it 1... –  bill s Feb 27 '14 at 20:17
Okay my mistake I should have done more research before asking the question. Is there a way to ask Mathematica to solve the equation for all of the variables? i.e. Solve[x+y==0,x] AND Solve[x+y==0,y] so that I get the solutions x==-y and y==-x as the two possibilities (I know this is a bad example because in this case it is easy to invert, but it won't always be easy to invert). –  Jeremy Upsal Feb 27 '14 at 22:50