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

Question

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

Answer 1

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.

Answer 2

This is not a formal approach but a trick I devised by myself. I shall be grateful if someone can provide it formally. Here's the approach.

Given, $$\left|z-r_{1}\right|+\left|z-r_{2}\right|+\left|z-r_{3}\right|+\left|z-r_{4}\right|+\left|z-r_{5}\right|+ \dots +\left|z-r_n\right|$$

Open the modulus such that the coefficient of $z$ is either $1$ or $0$. If it is zero, you straight away get the minimum, if it is $1$ just put $z$ such that it is the higher $r_i$ in the interval of expansion.

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Example: (solution to the given question)

I will assume $z$ between $r_2,r_3$. (One might do it for $r_3,r_4$ too). We have, $$z+\left(z-e^{ix}\right)+\left(5e^{ix}-z\right)+\left(7e^{ix}-z\right)+\left(32e^{ix}-z\right) = \left(-1+5+7+32\right)e^{ix}-z$$

Consequently $z=5e^{ix}$ (since it is limited between $[e^{ix},5e^{ix}]$) and hence the minimum $38e^{ix}$