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I would ask you to kindly provide me a way to solve this problem in mathematica way. Let the equation $z^2 + 9 -1.5 e^{- t z}=0$ where $z\in \mathbb C$ and $t\in \mathbb R^+$.

In order to plot graph of $t \mapsto \mathrm{Re}(z)$, I have triyed this:

f[z_] = z + 9  - 1.5 E^(- t z)

z = x + I y 

ExpToTrig[
  E^(-t (x + I y)) (-1.5 + 9. E^(t (x + I y)) + E^(t (x + I y)) x + 
    (0. + 1. I) E^(t (x + I y)) y)]
(0. + 0. I) - 1.5 Cosh[t x + I t y] + 9. Cosh[t x + I t y]^2 + 
  x Cosh[t x + I t y]^2 + (0. + 1. I) y Cosh[t x + I t y]^2 + 
  1.5 Sinh[t x + I t y] - 9. Sinh[t x + I t y]^2 - 
  x Sinh[t x + I t y]^2 - (0. + 1. I) y Sinh[t x + I t y]^2
TrigExpand[(0. + 0. I) - 1.5 Cosh[t x + I t y] + 
   9. Cosh[t x + I t y]^2 + 
   x Cosh[t x + I t y]^2 + (0. + 1. I) y Cosh[t x + I t y]^2 + 
   1.5 Sinh[t x + I t y] - 9. Sinh[t x + I t y]^2 - 
   x Sinh[t x + I t y]^2 - (0. + 1. I) y Sinh[t x + I t y]^2]
(9.` + 0.` I) + x + (0.` + 1.` I) y - 
  1.5` Cos[t y] Cosh[t x] + (0.` + 1.5` I) Cosh[t x] Sin[t y] + 
  1.5` Cos[t y] Sinh[t x] - (0.` + 1.5` I) Sin[t y] Sinh[t x]

In the last I can't do Real part. Any help is welcome.

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1  
It seems you have already accepted an answer despite posting a comment saying it does not answer your question. But if you really do want to plot $\operatorname{Re} z$ as a function of $t$, try sol = Solve[f[z] == 0, z]; Plot[Re[z] /. sol, {t, -1, 1}, PlotRange -> {-10, 0}] i.stack.imgur.com/wf7G8.png –  Rahul Narain May 27 at 17:34

1 Answer 1

up vote 1 down vote accepted

Assuming your goal is to plot t(z) where, z is real.

sol = Solve[z^2 + 9 - 3/2 E^(-t z) == 0, t];
usedSol = sol[[1]] /. {C[1] -> 0};
Plot[t /. usedSol, {z, 0, 1}]

enter image description here

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But I must to plot $t\mapsto \Re(z)$ where $z\in \mathbb C$ solution of the above equation. –  Zbigniew May 27 at 12:57

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