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?
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 $t_a=t_b$.
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.
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
$$ \sqrt(z) = \|z\|^{1/2} \, e^{i \arg(z) / 2} $$
For what purpose would it ever matter which root you choose? Any calculation that results in a real number will give you the same answer no matter which root you pick.
When you use the square root of $-1$ in an equation, do you worry which square root to pick? You might say you always pick $+i$, but I say the same thing--- and it's entirely possible that the root I call $i$ is the same as the root you call $-i$.
This causes no trouble because the real numbers you and I compute are always the same, regardless of whether our choices for $(-1)^{1/2}$ coincide --- and likewise for our choices of $(-1)^{1/4}$.
