Real solutions of the equation $\,z + e^{-z} - x = 0$ for $\,x > 1$ and $\,\Re(z) ≥ 0$ I have the complex equation $z + e^{-z} - x = 0$, where $z$ is complex and $x$ real, and I need to show that it has only one real solution for $x > 1$ and $\Re(z) \ge 0$. I do not see how this is possible, given that $z + e^{-z} = x$ should have all real numbers greater than $1$ as a possible solutions, as long as I keep plugging in real numbers for $z$. Am I misinterpreting what is being meant by a single solution? How can there be only one solution to this equation in right half of the complex plane?
 A: Mike is right: you are asked to show that for every $x>1$ there exists exactly one number $z$  that lies in the closed right halfplane $\overline{\mathbb H}$ and $z+e^{-z}-x=0$. 
Existence is based on the following observation: the function $t\mapsto t+e^{-t}$ increases from $1$ to $\infty$ as $t$ runs from $0$ to $\infty$. The intermediate value theorem implies that this function takes on the value $x$. 
Uniqueness is based on rewriting the equation as $x-e^{-z} =z$ and observing that the function $f(z) = x-e^{-z} $ is a weak contraction, meaning that 
$$|f(z)-f(w)|<|z-w|,\quad z,w\in\overline{\mathbb H}, \ z\ne w \tag{1}$$
Property (1) clearly rules out the possibility that $f(z)=z$ and $f(w)=w$ could hold for two distinct values $z,w$. 
To prove (1), write $f(z)-f(w) $ as an integral of derivative $f'(\zeta)=e^{-\zeta}$. Note that $|f'(\zeta)|\le 1$ for all $\zeta\in\overline{\mathbb H}$, with strict inequality in the open halfplane $\mathbb H$. This already implies (1) for all pairs $z,w\in\overline{\mathbb H}$ except for those where both $z$ and $w$ lie on the imaginary axis. For the remaining case one can argue as follows: pick $\theta$ such that $e^{i\theta}(f(z)-f(w))$ is real and positive; write
$$|f(z)-f(w)|= e^{i\theta}(f(z)-f(w)) \le \int_{z}^w   
\operatorname{Re}( e^{i\theta} f'(\zeta)) |d\zeta|$$
and observe that the integrand, being a trigonometric wave of amplitude $1$, is strictly less than one on a part of the interval of integration.
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{z + \expo{-z} - x = 0.\qquad\mbox{Let}\ z \equiv a + b\ic\,,\quad a, b \in {\mathbb R}.\qquad x > 1\,,\ a > 0}$

$$
a + \expo{-a}\cos\pars{b} - x = 0\,,\qquad b - \expo{-a}\sin\pars{b} = 0
\quad\imp\quad
\left\vert%
\begin{array}{rcl}
\cos\pars{b} & = & \pars{x - a}\expo{a}
\\
\sin\pars{b} & = & b\expo{a}
\end{array}\right.
$$

It leads to $\pars{x - a}^{2}\expo{2a} + b^{2}\expo{2a} = 1$ or/and
$\pars{x - a}^{2} + b^{2} = \expo{-2a} < 1$. Then, $\verts{x - a} < 1$ and
$\verts{b} < 1$:
$$
0 < x - 1 < a < x + 1\quad\mbox{and}\quad -1 < b < 1
$$
Also, if $a + b\ic$ is a solution, $a - b\ic$ is a solution too: $a \pm \verts{b}\ic$ are $\large\tt 2$ solutions ( at least ).
