Find minimum value of $|z|+|z-e^{ix}|+|z-5e^{ix}|+|z-7e^{ix}|+|z-32e^{ix}|$ where $z$ is a complex number 
Find minimum value of $$|z|+|z-e^{ix}|+|z-5e^{ix}|+|z-7e^{ix}|+|z-32e^{ix}|$$ where $z$ is a complex number.

we notice that we have to minimze the sum of distances from the points in the argand plane  $(0,e^{ix},5e^{ix},7e^{ix},32e^{ix})$.We also see these points are on a straight line which makes an angle $x$ with real axix. .We now rotate the plane by $x$ radians anticlockwise so that the points lie on the real line.Let the coordinates of $z$  after rotation be $(h,k)$.Intuitively this does not change the minimum.So now we have to minimise: $$\sqrt{h^2+k^2}+\sqrt{{(h-1)}^2+k^2}+\sqrt{{(h-5)}^2+k^2}+\sqrt{{h-7)}^2+k^2}+\sqrt{{(h-32)}^2+k^2}$$  which happens when
$k=0$.We minimize  $$f(h)=|h|+|h-1|+|h-5|+|h-7|+|h-32|$$
But $f(h)$ becomes  a lenear function for each of the intervals $h\in [-\infty,0],[0,1],[1,5],[5,7],[7,32],[32,\infty]$.But a lenear function achieves its maximum and minimum at the endpoints of interval.
We seek:$$\min(f(0),f(1),f(5),f(7),f(32))=f(5)=38$$
Is it good?
I am looking for alternative proofs

You can use only the triangle inequality:
$$|z| + |z-32 e^{ix}| \ge |32 e^{ix}| = 32 \\
|z-e^{ix}| + |z - 7 e^{ix}| \ge |7 e^{ix} - e^{ix}| = 6 \\
|z- 5e^{ix}| \ge 0$$
Adding up we get : $f(z) \ge 32 + 6 + 0 = 38$.  We have equality for $z = 5e^{ix}$  ( an easy but important check). Conversely, if we have equality then necessarily $|z- 5 e^{ix}| = 0$ so $z = 5 e^{ix}$.
Note: $x$ is a parameter, $z$ is the variable.
