Prove that for every \(\lambda>1\), the equation \[ ze^{\lambda-z}=1 \] has exactly one root in the disk \(|z|<1\), and that this root is real.
Solution
Define \[ G(z)\coloneqq z-e^{z-\lambda}. \] The equation in the problem is equivalent to \(G(z)=0\).
1. The function \(G\) has exactly one zero in the disk \(\abs{z}<1\), counted with multiplicity.
On the circle \(\abs{z}=1\), \[ \abs{e^{z-\lambda}} = e^{\operatorname{Re}z-\lambda} \leq e^{1-\lambda} <1 = \abs{z}, \] because \(\lambda>1\). Hence, by Rouché’s theorem, the functions \[ z \qquad\text{and}\qquad z-e^{z-\lambda}=G(z) \] have the same number of zeros in \(\abs{z}<1\), counted with multiplicity. The function \(z\) has exactly one such zero, namely \(0\) with multiplicity one. Therefore \(G\) also has exactly one zero in the disk.
2. The equation \(G(x)=0\) has a real solution \(x\in(0,1)\).
For real \(x\), \[ G(x)=x-e^{x-\lambda} \] is continuous. Moreover, \[ G(0)=-e^{-\lambda}<0 \] and \[ G(1)=1-e^{1-\lambda}>0, \] since \(\lambda>1\). The intermediate value theorem therefore gives some \(x\in(0,1)\) with \(G(x)=0\).
3. The equation\[ ze^{\lambda-z}=1 \]has exactly one root in \(\abs{z}<1\), and that root is real.
The equations \[ ze^{\lambda-z}=1 \qquad\Longleftrightarrow\qquad z=e^{z-\lambda} \qquad\Longleftrightarrow\qquad G(z)=0 \] are equivalent. Step 1 gives exactly one zero of \(G\) in the disk, while step 2 exhibits a real zero there. Hence the unique root is real.
4. Q.E.D.
Step 3 is the required conclusion.