= Solution
The <quadratic formula> gives
$$
z=-\frac b2\pm\frac i2\sqrt{4-b^2},
$$
so both <roots of a polynomial> have <real part> $-b/2$. Writing $z=x+iy$, the <complex exponential function> satisfies $e^z=e^x(\cos y+i\sin y)$, and hence its <modulus> is $e^x$. Therefore
$$
|e^z|=e^{-b/2}<1
\quad\Longleftrightarrow\quad b>0.
$$
Intersecting this condition with the allowed parameter interval gives the answer \b[for either root]:
$$
\boxed{0<b\le2.}
$$
The endpoint $b=2$ is included: the repeated <root of a polynomial> is $z=-1$, whose exponential has <modulus> $e^{-1}<1$.
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