Let \(0 \leq t < 1\). Show that \(f(z) = e^{tz} + z + 1\) has exactly one zero (counting multiplicities) in the left half-plane \(\{z : \operatorname{Re} z < 0\}\).
Hint: Consider first \(|z| < \varepsilon\) for \(\varepsilon > 0\).
The official Fall 2022 source states only \(t<1\). That is false for negative \(t\): for \(t\ne0\), the equation is equivalent to \[ -t(z+1)=W_k(te^{-t}), \] and for negative \(t\) infinitely many branches give zeros in the left half-plane. The hypothesis above records the intended nonnegative range.
Solution
Fix \(\varepsilon>0\) small enough that \[ e^{t\varepsilon}<1+\varepsilon. \] Such an \(\varepsilon\) exists because \(t<1\) and both sides equal \(1\) at \(\varepsilon=0\), while their derivatives there are \(t\) and \(1\).
Consider the half-plane \(\operatorname{Re}z<\varepsilon\). On its boundary \(z=\varepsilon+iy\), \[ |e^{tz}|=e^{t\varepsilon}<1+\varepsilon\le|z+1|. \] Since \(t\ge0\), the exponential is bounded by \(e^{t\varepsilon}\) throughout that half-plane. On a sufficiently large left semicircle closing the boundary, \(|z+1|\) is therefore larger than \(|e^{tz}|\) as well. Rouché’s theorem on the resulting half-disk shows that \[ e^{tz}+z+1 \] and \(z+1\) have the same number of zeros in \(\operatorname{Re}z<\varepsilon\), namely one, counting multiplicity.
It remains to show that this zero is actually in the left half-plane. If \(z=iy\) lies on the imaginary axis and is a zero, then \[ |1+iy|=|e^{ity}|=1, \] which forces \(y=0\); but \(f(0)=2\). Hence there are no zeros on the imaginary axis. Since the unique zero in \(\operatorname{Re}z<\varepsilon\) cannot lie in \(0\le\operatorname{Re}z<\varepsilon\) for all sufficiently small \(\varepsilon\), it lies in \(\operatorname{Re}z<0\). Thus there is exactly one left-half-plane zero, counted with multiplicity.