Fall 2011 - Problem 8

Solution.

Notice that any solution satisfies $z = 2 + e^{-z}$. Thus, $\abs{z} \leq 2 + \abs{e^{-z}} \leq 3$. Hence, the zeroes are contained in the half-disk $U = \set{z \in \C \mid \abs{z} < 3, \Re\p{z} > 0}$. Set $f\p{z} = z - 2$ and $g\p{z} = -e^{-z}$. Observe that if $\Re\p{z} = 0$, then $z = ai$ for some $a \in \R$ and

Similarly, if $\abs{z} = 3$,

Thus, $\abs{g\p{z}} < \abs{f\p{z}}$ on $\partial{C}$, so by Rouche's theorem, $f\p{z} = z - 2$ and $f\p{z} + g\p{z} = z - 2 - e^{-z}$ have the exact number of roots in $C$. Clearly $f\p{z}$ has $z = 2$ as its only root in $C$, so $z - 2 - e^{-z} = 0$ has precisely one root in $C$ and hence in $H$.