This is just a quick shot, so I may have made a mistake, but I think the answer is that $z_1=1$ and $z_2=0$ is an optimum, giving a result of $2$.
Since there is symmetry with respect to rotation, fix $z_1$ to be real.
Since there is symmetric treatment of $z_1$ and $z_2$, I suspect that either one of them is zero or else their magnitudes are equal. This isn't certain --- it's also possible that the optimum is two unequal, nonzero magnitudes, and you can arbitrarily choose which one you want to be the larger.
Assuming equal magnitudes, we have $z_1=1/\sqrt{2}$ and $z_2=e^{i\phi}/\sqrt{2}$. The minimum result then occurs when $\phi=0$, giving the claimed result.
Assuming that one has magnitude 0, we also get a result of 2.
To be certain of the result, you really need to take $z_1=\cos\theta$, $z_2=\sin\theta e^{i\phi}$, and minimize simultaneously with respect to $\theta$ and $\phi$.
EDIT: Here's a sketch for a complete proof. The function to be minimized is
$f=\sqrt{2}\sin(\theta+\pi/4)\left(\sqrt{1+\sin2\theta\cos\phi}+\sqrt{1-\sin2\theta\cos\phi}\right)$
where $0 \le \theta \le \pi/2$ and $0 \le \phi \le \pi/2$. (The expression is invalid for $\theta$ in the second quadrant, and points with $\phi$ in the second quadrant give the same result as points in the first quadrant.)
It should be pretty easy to prove that on this rectangle in the $(\theta,\phi)$ plane, $\partial f/\partial\phi \ge 0$, so you only have to search for minima along the edge at $\phi=0$. Along this edge, there are no local minima at differentiable points, but there is a minimum at the nondifferentiable point $\phi=\pi/4$, as well as at the corners. All three of these minima give $f=2$.