Is a complex Keplerian eccentricity $e=\sqrt{1+\frac{2\varepsilon h^2}{\mu^2}}$ possible? Wikipedia provides the following formula for the Keplerian orbital eccentricity:
$$e=\sqrt{1+\frac{2\varepsilon h^2}{\mu^2}}$$
Considering that a bound object means negative energy, if the magnitude of the ratio in the square root is bigger than 1, then the eccentricity turns out to be complex.
So, what am I missing?
 A: No, this isn't possible, basically because the specific energy is bounded below by the centrifugal barrier produced by the specific orbital angular momentum when it combines with the gravitational potential. 
The formula you quote involves three main ingredients in calculating the eccentricity:


*

*The specific orbital energy, $\varepsilon = \frac{v^2}{2} - \frac{\mu}{r}$.

*The standard gravitational parameter, $\mu=GM$, which is basically a re-normalization of the larger body's mass.

*The specific relative angular momentum $h=L/m$.
At any given point during the orbit, the velocity can be decomposed into a radial component $v_r$ and an azimuthal component $v_\phi$, and it is only the latter that contributes to the angular momentum, so that
$$
h = L/m = rp_\phi/m =  r v_\phi
$$
at all times, but both components contribute to the kinetic energy,
$$
\frac12 v^2 = \frac12 v_r^2 + \frac12 v_\phi^2.
$$
Here the second term can be re-cast in terms of the angular momentum and the distance, giving rise to the so-called centrifugal barrier for the dynamics, in which the specific energy reads
\begin{align}
\varepsilon 
& = \frac12 v_r^2 + \frac12 v_\phi^2 - \frac{\mu}{r}
\\ & = \frac12 v_r^2 + \frac12 \frac{h^2}{r^2} - \frac{\mu}{r}
\\ & \geq \frac12 \frac{h^2}{r^2} - \frac{\mu}{r},
\end{align}
because the radial kinetic energy term $\frac12 v_r^2$ is nonnegative. 
This means that the specific energy must be no smaller than $f(r) = \frac12 \frac{h^2}{r^2} - \frac{\mu}{r}$, i.e. that it must be no more negative than $-\left| \frac12 \frac{h^2}{r^2} - \frac{\mu}{r} \right|$. As it turns out, this is a simple fraction of $r$ which (so long as $h\neq 0$) always has a minimum at some radius $r=r_*$. To find this minimum, you simply set the derivative of $f(r)$ to zero,
\begin{align}
0 & = f'(r_*)
\\ & = -2 \frac12 \frac{h^2}{r_*^3} + \frac{\mu}{r_*^2}
\end{align}
and you solve, giving trivially $r_* = \frac{h^2}{\mu}$. If you put this back into the above inequality, you get
\begin{align}
\varepsilon 
& \geq f(r) \geq f(r_*)
 = \frac12 \frac{h^2\mu^2}{h^4} - \frac{\mu^2}{h^2}
 = -\frac{\mu^2}{2h^2},
\end{align}
and if you re-phrase this a bit, you get that
$$
\frac{2\varepsilon h^2}{\mu^2} + 1 \geq 0
$$
for all possible orbits, which rules out any complex-valued eccentricities.
