In a wave equation, which is not ordinary kinematics, we can, indeed, have an imaginary wave speed: given the ordinary wave equation, taken in one dimension for simplicity,
$$\frac{\partial^2 u}{\partial t^2} = v_w^2 \frac{\partial^2 u}{\partial x^2}$$
if $v_w = iv$ for $v$ real, then $v_w^2 = -v^2$ and the WE looks like
$$\frac{\partial^2 u}{\partial t^2} = -v^2 \frac{\partial^2 u}{\partial x^2}$$
To understand the meaning, note that a wave equation has elementary solution
$$u(x, t) = A \cdot \sin\left(kx - \omega t + \phi\right)$$
where the wave number $k$ and angular frequency $\omega$ are required to satisfy
$$\frac{\omega}{k} = v_w$$
If $v_w$ is imaginary, then that means one of $\omega$ or $k$ will be so, and you get sine of a complex number, which is also in general a complex number. This will, then, only make sense if the value of the wave, $u(x, t)$, itself can make sense as a complex number, too. But it will not involve motion into imaginary directions of space, because the domain of variable $x$, the spatial position, is still the real numbers.
ADD 2: I think this should have a real solution, actually - it's a real differential equation after all...
ADD 3: Yes ... I had a hunch but had to work it out, the relevant real-number solution is ... WEEERD! Don't forget the angle sum formula, and you can get, writing as $kx - \omega t + \phi = (kx + \phi) - \omega t$, using the identity that $\cos(ix) = \cosh(x)$ with imaginary $\omega$ as per the relation above,
$$u(x, t) = A\ \sin(kx + \phi)\ \cosh(\omicron t) = [A \cosh(\omicron t)] \sin(kx + \phi)$$
and throw away the imaginary part (justifiable as being a linear combination with a conjugate) as a working real solution, where I've used $\omicron$ to denote $\omega = i\omicron$. Note that two derivatives in $x$ gives you $-k^2$ out front but two derivatives in $t$ gives you $\omicron^2$ out front, so the desired negative sign appears.
Hence, imaginary wave speed means the wave pattern is sitting still but growing exponentially in amplitude!
ADD 4: I believe also there may be a way to get exponential decay, as well.