Solution
= Solution
For $z=x+iy$, the <complex exponential> has <complex modulus> $e^x$. Equality to the positive real number $e$ therefore forces $x=1$ and $\cos y+i\sin y=1$. The latter holds precisely when $y$ is an integer multiple of $2\pi$. Hence \b[all solutions are]
$$
\boxed{z=1+2\pi i k,\qquad k\in\mathbb Z.}
$$
Substitution verifies every listed value; the modulus and angle arguments exclude any others.