Discontinuity in the angle of a complex exponential signal
That's not really a discontinuity. On a circle the two points $-\pi$ and $+\pi$ are identified: They are the same point. That is true for all $x$ and $x+n 2\pi$ for integer $n$.

If you would like to keep the continuity obvious, you might want to switch to a phasor representation of angles. Those are complex numbers on the unit circle $\exp(i\phi)$ and they form a group under multiplication. They nicely absorb the non-obvious topology of the phase angle.