How to minimize the following function?

$$f(x)=\max\left(\|x-ae^{i\theta/2}\|,\|x-be^{i\theta/2}\|\right) $$ hence by assuming $a>0$ we are in the following configuration:

Now consider that

1. The distance from a point is a convex function;
2. If $f,g$ are convex functions, $\max(f,g)$ is a convex function;
3. Convex functions over convex domains attain their maxima at the boundary;
4. The locus of points $P$ such that $\max(PA,PB)=d$, if non-empty, is given by the union of two circle arcs, symmetric with respect to the perpendicular bisector of $AB$;
5. Let $A=ae^{i\theta/2}$ and $B=be^{i\theta/2}$. The previous points ensure that the solution of the minimization problem is given by the the projection of $B$ on the $ab$ line, if this point belongs to the $ab$ segment. Otherwise the solution is given by one of the endpoints of $ab$. This is the only part in which $\theta$ plays an active role.

I will let you finish.

Hint:

The two arguments are equal when

$$x^2-2ax\cos\phi+a^2=x^2-2bx\cos\phi+b^2$$ or

$$\hat x=\frac{a+b}{2\cos\phi}$$ where $0<\cos\phi<1$. With $b>a$, $\hat x>a$ and $\hat x$ reaches $b$ when $\cos\phi=\dfrac{a+b}{2b}$.

So two cases:

• if $\cos\phi\le\dfrac{a+b}{2b}$, you minimize the second argument over $[a,b]$,

• if $\cos\phi>\dfrac{a+b}{2b}$, you minimize the second argument over $[a,\hat x]$, and the first argument over $[\hat x,b]$.

The local minima of these expressions are achieved when $\bar x=a\cos\phi$ (resp. $\bar x=b\cos\phi$). The first $\bar x$ is always out of range, and the second also if $b\cos\phi<a$.

Finally, you have to compute the values at the relevant interval endpoints and the local maxima, depending on the existence of $\hat x$ and $\bar x$.