= Solution
Write the <complex number> as $z=x+iy$. Since $iz=-y+ix$, the <modulus of the complex exponential> gives
$$
|e^{iz}|=e^{-y}.
$$
This equals one precisely when $y=0$. The <quadratic formula> shows that the <imaginary part> of either root is $\pm\sqrt{4-b^2}/2$. Consequently the roots are real exactly at the two endpoints:
$$
\boxed{b=-2\ \text{or}\ b=2.}
$$
\b[Both endpoints satisfy the condition.] The repeated roots are respectively $1$ and $-1$; their multiples by $i$ have exponentials on the <unit circle>.
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