Real solutions of the equation $\,z + e^{-z} - x = 0$ for $\,x > 1$ and $\,\Re(z) ≥ 0$

Question

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?

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{z + \expo{-z} - x = 0.\qquad\mbox{Let}\ z \equiv a + b\ic\,,\quad a, b \in {\mathbb R}.\qquad x > 1\,,\ a > 0}$

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

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 ).