Root of $i$, which one to take? The propagator of a free particle in 1d is 
$$ K(x_b, t_b; x_a, t_a ) = \sqrt{\frac{m}{2\pi i \hbar (t_b-t_a)}}  \exp \left [ \frac{i m (x_b-x_a)^2}{2 \hbar (t_b-t_a)} \quad \right ] .$$ It looks nice. 
But, here we have a square root of $i$. Between the two roots, which one should be taken? Based on what rule?
 A: That propagator is nothing but the analytic continuation of the Green function of heat equation from real positive to imaginary values of $t_b-t_a$.  The cut in the complex plane to make single valued the square root has therefore to be put along the negative real axis, or however in the semiplane $x<0$. With this cut the square root is well defined.
All that means that $\sqrt{i} = e^{i\pi/4}$ is the mathematically correct choice in that formula, the one producing a Dirac delta for $t_a=t_b$.
A: Define $\Delta t := t_b-t_a$ and $\Delta x := x_b-x_a$. 
One should ensure that 
$$\tag{1}{\rm Re}(i\Delta t)~>~0$$ 
is positive in order for the exponential factor 
$$\tag{2}\exp \left [- \frac{ m}{2 \hbar}\frac{(\Delta x)^2}{ i\Delta t} \right]$$ 
to be exponentially damped. 
Equivalently, one should perform the Feynman $i\epsilon$ prescription, i.e., substitute $ \Delta t\to\Delta t-i\epsilon$ in the propagator. This requirement (1) is to ensure that 
$$\tag{3} \langle x_b ,t_b | x_a ,t_a \rangle ~\longrightarrow~\delta(\Delta x) \quad \text{for} \quad  \Delta t \to 0$$
when one picks the branch of the square root that has positive real part.
A: Since the propagator is defined by the relation
$$
\psi(x,t) = \int K(x',x,t,t') \psi(x',t') dx'\, dt'
$$
A sing ambiguity would result in a change of phase of the wavefuntion, which does not alter the predictions of quantum mechanics. So it doesn't matter which square root you take, as a matter of fix ideas i always take the root
$$
\sqrt(z) = \|z\|^{1/2} \, e^{i \arg(z) / 2}
$$
