Solve an equation involving Exp over the complexes

Question

I need some help to solve an exponential equation with Mathematica over the complexes. This is the equation I need to solve. I've already used FullSimplify.

e^{4-2x-2e^{2-2x}x}x - x =0

After defining the whole function as G[x], I used the command

rts = Reduce[G[x] == 0, x, Complexes]

but the message is

Reduce::nsmet: This system cannot be solved with the methods available to Reduce.

I also tryed to use Solve and FindRoot but it doesn't work.

Can you help me please?

Answer 1

NSolve does the job by

NSolve[E^(4 - 2 x - 2 E^(2 - 2 x) x) x - x == 0 &&  Abs[x] <= 10, x, Complexes];Dimensions[%]

{279, 1}

Addition. It's strange that

NSolve[E^(4 - 2 x - 2 E^(2 - 2 x) x) x - x == 0 && Abs[x] <= 10, x,  Complexes, WorkingPrecision -> 30];Dimesions[%]

{268, 1}

and

NSolve[E^(4 - 2 x - 2 E^(2 - 2 x) x) x - x == 0 && Re[x] <= 10 && 
Re[x] >= -10 && Im[x] <= 10 && Im[x] >= -10, x, Complexes, 
WorkingPrecision -> 30];Dimensions[%]

{268, 1}

I'll try to verify it with Maple.

Addition 2. The verification

Table[E^(4 - 2 x - 2 E^(2 - 2 x) x) x - x /. sol[[j]], {j, 1, 268}]

confirms all 268 solutions.

Answer 2

What about x=0, x=1? Any other solution has exponent 0, too, so solve

2/x-1-e^(2-2x) == 2/x-1 - (e/e^x)^2 ==0

The other roots seem to sit on $x =1 + i y$

  FindRoot[2/x - 1 - E^2/E^(2 x), {x, 1 + 5 I}]

    {x -> 1. + 4.49341 I}

2/(1 + I x) - 1 - E^2/E^(2 (1 + I x)) == 0 // FullSimplify

  (x Cos[x] - Sin[x])/(-I + x) == 0


     Plot[x Cos[x] - Sin[x], {x, -120, 120}]

[

The ComplexPlot3D[1/f] exhibits the structure of zeros as poles

and a finer resolution