# Plot a real part of complex equation depending on a parameter

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.

-
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 May 27 '14 at 17:34

sol = Solve[z^2 + 9 - 3/2 E^(-t z) == 0, t];

But I must to plot $t\mapsto \Re(z)$ where $z\in \mathbb C$ solution of the above equation. –  Zbigniew May 27 '14 at 12:57