Let \(1<\lambda<\infty\) and \[ f_\lambda(z)=z+\lambda-e^z. \] Prove that \(f_\lambda\) has exactly one zero in the half-plane \(\operatorname{Re}z<0\), and that this zero is real.
Solution
1. Every zero of \(f_\lambda\) in the half-plane\[ \operatorname{Re}z<0 \]is real.
Suppose \[ z=x+iy, \qquad x<0, \] and \(f_\lambda(z)=0\). Then \[ z+\lambda=e^z=e^x(\cos y+i\sin y). \] Taking imaginary parts gives \[ y=e^x\sin y. \] If \(y\ne0\), then \[ |y| = e^x|\sin y| \le e^x|y| < |y|, \] because \(x<0\) implies \(e^x<1\). This is impossible. Hence \(y=0\), so \(z\) is real.
2. The real function\[ g(x)\coloneqq x+\lambda-e^x \]has exactly one zero on \((-\infty,0)\).
For \(x<0\), \[ g'(x)=1-e^x>0, \] so \(g\) is strictly increasing on \((-\infty,0)\). Moreover, \[ \lim_{x\to-\infty}g(x)=-\infty \] because \(e^x\to0\), while \[ g(0)=\lambda-1>0. \] The intermediate value theorem therefore gives a zero \[ x_\lambda\in(-\infty,0), \] and strict monotonicity makes it unique.
3. The number \(x_\lambda\) from step 2 is the unique zero of\(f_\lambda\) in the half-plane \(\operatorname{Re}z<0\).
Since \(x_\lambda\) is real, \[ f_\lambda(x_\lambda)=g(x_\lambda)=0, \] so a zero in the required half-plane exists. By step 1, every zero in that half-plane is real, and step 2 shows that the only possible real zero there is \(x_\lambda\).
4. Q.E.D.
Step 3 proves both uniqueness and reality of the zero in \(\operatorname{Re}z<0\).