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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?

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  • $\begingroup$ Should be G[x_] := E^(4 - 2 x - 2 E^(2 - 2 x) x) x - x. $\endgroup$
    – cvgmt
    Oct 1, 2023 at 10:45
  • $\begingroup$ @cvgmt I've corrected it, but it doesn't work $\endgroup$
    – Marthy92
    Oct 1, 2023 at 10:54
  • 1
    $\begingroup$ isn't $x=0$ a solution? $\endgroup$
    – Nasser
    Oct 1, 2023 at 11:15

2 Answers 2

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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.

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  • $\begingroup$ I switched to real variables. The command of Maple DirectSearch:-SolveEquations([exp(4-2*x-2*exp(2-2*x)*cos(2*y)*x-2*exp(2-2*x)*sin(2*y)*y)*cos(-2*y+2*exp(2-2*x)*sin(2*y)*x-2*exp(2-2*x)*cos(2*y)*y)*x-exp(4-2*x-2*exp(2-2*x)*cos(2*y)*x-2*exp(2-2*x)*sin(2*y)*y)*sin(-2*y+2*exp(2-2*x)*sin(2*y)*x-2*exp(2-2*x)*cos(2*y)*y)*y-x = 0, exp(4-2*x-2*exp(2-2*x)*cos(2*y)*x-2*exp(2-2*x)*sin(2*y)*y)*sin(-2*y+2*exp(2-2*x)*sin(2*y)*x-2*exp(2-2*x)*cos(2*y)*y)*x+exp(4-2*x-2*exp(2-2*x)*cos(2*y)*x-2*exp(2-2*x)*sin(2*y)*y)*cos(-2*y+2*exp(2-2*x)*sin(2*y)*x-2*exp(2-2*x)*cos(2*y)*y)*y-y = 0], $\endgroup$
    – user64494
    Oct 1, 2023 at 17:21
  • $\begingroup$ {x >= -10, y >= -10, x <= 10, y <= 10}, AllSolutions, solutions = 100, tolerances = 10^(-11), number = 1200); finds 52 solutions. $\endgroup$
    – user64494
    Oct 1, 2023 at 17:22
  • $\begingroup$ Test for solutions near Re =1: Select[solm, (0.999999 < Re[#] < 1.00001 &)] // Length ->71 , control by Plot[Re[E^(4 - 2 x - 2 E^(2 - 2 x) x) x - x], {x, 0.9999999, 1.00000001}] showing bit noise near 0. x=1 is a 3. order zero. $\endgroup$
    – Roland F
    Oct 2, 2023 at 8:18
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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}]

[x Cos[x] - Sinx

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

complex plot 1

and a finer resolution

complex plot 2

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  • $\begingroup$ Here are several results of NSolve: {x -> -2.17747124047226705535721254829 + 1.98532585029532272308446021163 I}, {x -> -1.44292479059622017211280337669 - 0.59744381618830898847336446238 I}, {x -> -1.44292479059622017211280337669 + 0.59744381618830898847336446238 I}. $\endgroup$
    – user64494
    Oct 1, 2023 at 13:07

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