# Solve an equation involving Exp over the complexes

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.

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

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}

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.

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


confirms all 268 solutions.

• 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], Commented Oct 1, 2023 at 17:21
• {x >= -10, y >= -10, x <= 10, y <= 10}, AllSolutions, solutions = 100, tolerances = 10^(-11), number = 1200); finds 52 solutions. Commented Oct 1, 2023 at 17:22
• 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. Commented Oct 2, 2023 at 8:18

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

• Here are several results of NSolve: {x -> -2.17747124047226705535721254829 + 1.98532585029532272308446021163 I}, {x -> -1.44292479059622017211280337669 - 0.59744381618830898847336446238 I}, {x -> -1.44292479059622017211280337669 + 0.59744381618830898847336446238 I}. Commented Oct 1, 2023 at 13:07