For `z=x+iy, ` the equation becomes
`12 ( x^2 + y^2 ) = 2 ( x + 2 )^2 + 2y^2 + (x^2 - y^2 + 1 )^2 + 4x^2y^2 + 31 .`
Open parentheses and transform it to
`( x^2 + y^2 )^2 + 42 = 2 ( 2x - 1 )^2 + 12y^2 .`
Consider `x = -1 , ` then we get
`( 1 + y^2 )^2 + 42 = 18 +12y^2 , ` `y^4 - 10y^2 + 25 = 0 , ` ` y^2 = 5 , ` `y = +- sqrt ( 5 ) .`
The strange fact is that there are no more solutions, so `z = -1 +- i sqrt ( 5 ) ` and
`z + 6 / z = -1 +- i sqrt ( 5 ) + 6 / 6 ( -1 -+ i sqrt ( 5 ) ) = -2 .`
To prove that there are no more solutions, consider
`f(x,y)=( x^2 + y^2 )^2 + 42 - 2 ( 2x - 1 )^2 - 12y^2`
and found its critical points:
`(del f) / (del x) = 4x(x^2+y^2)-8(2x-1), ` `(del f)/(del y)=4y(x^2+y^2)-24y.`
If both of them are zero, two possibilities arise. One is `x^2+y^2=6 ` from the second equation and `x = -1 ` from the second, which gives the already known points. Another is `y=0 ` from the second equation and `x^3-4x+2=0 ` from the first.
It is possible to prove that our solutions are local minima; is is also simple to prove that `f(x,0)gt0. ` Because for large `x, y ` `f ` is also positive, we can conclude that the points we found are points of the global minima, so there are no more solutions.