How would I go about minimizing the expression

$\left(|z_1| + |z_2|\right) \times \left(|z_1 + z_2| + |z_1 - z_2|\right)$

subject to the constraint

$|z_1|^2 + |z_2|^2 = 1$

given that $z_1$ and $z_2$ can be complex numbers?

I thought of trying Lagrange multipliers, but it doesn't seem possible because there are an infinite number of solutions (and solving a 5-equation, 5-variable system is a bit painful).

Any hints on how I could do this?

  • $\begingroup$ There is a large literature on this sort of thing. Terms to google include linear programming, nonlinear programming, and convex optimization. The function you're trying to optimize is convex, which probably helps, but it's not analytic, which makes it harder. The constraint is nonlinear and not convex, which makes it harder. The whole problem is symmetric with respect to rotation in the complex plane, so you can reduce it to three degrees of freedom by taking $\Im(z_1)=0$ if you wish. It's also symmetric under interchange of $z_1$ and $z_2$, so I suspect you have $|z_1|=0$, $1/\sqrt{2}$,or 1. $\endgroup$ – Ben Crowell Jan 23 '12 at 21:18
  • $\begingroup$ Whoa I didn't know any of those terms -- thanks a lot for pointing that out! $\endgroup$ – Mehrdad Jan 23 '12 at 22:19

EDIT: my first try found the maximum rather than the minimum.

Suppose wlog $|z_1| \ge |z_2|$. Fix $z_1$ and $|z_2|$ for the moment, and write $z_2 = w z_1$. Then we want to minimize $|z_1 + z_2| + |z_1 - z_2| = |z_1| (|w + 1| + |w - 1|)$ subject to $|w| = |z_2|/|z_1|$. Geometrically, $|w + 1| + |w - 1|$ is the sum of the distances from $w$ to $1$ and $-1$, which is $2$ on the line segment $[-1,1]$ and greater than $2$ otherwise, and $w$ is on a circle centred at $0$ with radius less than $1$, so the minimum value is attained at $w = \pm |z_2|/|z_1|$. Now write $|z_1| = \cos \theta$ and $|z_2| = \sin \theta$, $0 \le \theta \le \pi/4$. We have $|z_1 + z_2| + |z_1 - z_2| = 2 \cos \theta$, so the objective is to minimize $2 (\cos \theta + \sin \theta)\cos \theta$ for $0 \le \theta \le \pi/4$, and it is easy to see that the minimum value is $0$, attained at both $\theta=0$ and $\theta=\pi/4$.

  • 2
    $\begingroup$ I get the same answer for the maximum, but he wants the minimum. $\endgroup$ – Ben Crowell Jan 23 '12 at 23:10
  • 1
    $\begingroup$ Thanks, fixed it. I still think that the maximum is a better problem than the minimum. $\endgroup$ – Robert Israel Jan 24 '12 at 0:27
  • $\begingroup$ Yes, the geometrical approach to the maximum problem looked sweet from what I was able to glean while it was still up. $\endgroup$ – Ben Crowell Jan 24 '12 at 0:30

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


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 $f=2$.

  • $\begingroup$ I can definitely guess the answer the same way (assuming min/max or equal), but the trouble I'm having is proving that this is the answer. Also I don't understand why setting $z_1 = \cos \theta$ and $z_2 = \sin \theta e^{i \phi}$ is allowed -- sure, we can set $z_1$ to be real, but how can you assume that $z_2$ only depends on 1 independent variable ($\phi$)? Doesn't it depend on 2 independent variables (excluding $\theta$) since its real and imaginary components are independent? $\endgroup$ – Mehrdad Jan 23 '12 at 22:18
  • $\begingroup$ Or to say it a different way -- the original expression has four semi-independent variables (2 for each of 2 complex variables), but your example has only two. I can imagine maybe one of them going away, but where did the other go? $\endgroup$ – Mehrdad Jan 23 '12 at 22:37
  • $\begingroup$ One degree of freedom goes away because of symmetry with respect to rotation in the complex plane. One goes away because of your constraint. That leaves only two, which I've parametrized as $\theta$ and $\phi$. The $\sin\theta$ and $\cos\theta$ stuff is to make your constraint automatically satisfied. $\endgroup$ – Ben Crowell Jan 23 '12 at 23:13

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