For each \(\lambda>1\), prove that the equation \[ z=\lambda-e^{-z} \] has exactly one root in the half-plane \(\operatorname{Re}z\ge0\), and that this root is real.
Solution
Write \[ z=x+iy, \qquad x,y\in\RR. \]
1. If \(z\) is a root with \(x\ge0\), then\[ y=e^{-x}\sin y. \]
The equation \[ x+iy = \lambda-e^{-x}e^{-iy} \] becomes \[ x+iy = \lambda-e^{-x}\cos y +i e^{-x}\sin y. \] Equality of imaginary parts gives the displayed identity.
2. Every root in the half-plane \(\operatorname{Re}z\ge0\) isreal.
Suppose \(y\neq0\). By step 1 and \(x\ge0\), \[ \abs{y} = e^{-x}\abs{\sin y} \le \abs{\sin y}. \] For every nonzero real \(y\), \[ \abs{\sin y}<\abs{y}. \] Combining these two inequalities gives the contradiction \[ \abs{y}<\abs{y}. \] Hence \(y=0\).
3. A real number \(x\ge0\) is a root exactly when\[ h(x) \coloneqq x+e^{-x}-\lambda = 0. \]
With \(y=0\), the original equation is \[ x=\lambda-e^{-x}, \] which is equivalent to the displayed equation.
4. The function \(h\) has at least one zero in \((0,\lambda)\).
Since \(\lambda>1\), \[ h(0)=1-\lambda<0, \] while \[ h(\lambda)=e^{-\lambda}>0. \] The intermediate value theorem therefore gives a zero in \((0,\lambda)\).
5. The function \(h\) has at most one zero in \([0,\infty)\).
For \(x>0\), \[ h'(x)=1-e^{-x}>0. \] Thus \(h\) is strictly increasing on \((0,\infty)\). Since \(h(0)<0\), it can have at most one zero in \([0,\infty)\).
6. Consequently, the original equation has\[ \boxed{\text{exactly one root in $\operatorname{Re}z\ge0$, and it is real}.} \]
Step 2 shows that every root in the closed right half-plane is real. Steps 4 and 5 show that the corresponding real equation has exactly one nonnegative root.
7. Q.E.D.
Step 6 is the required conclusion.