Minimum value of $|z+1|+|z-1|+|z-i|$

Question

How to find the minimum value of $|z+1|+|z-1|+|z-i|$.

I have tried geometrically etc but failed.

Answer 1

let $z=a+bi$ then we have$$f(a,b)=\sqrt{(a+1)^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}$$ with the help of calculus we get $$f(a,b)\geq 1+\sqrt{3}$$ the equal sign holds if $a=0$ and $b=\frac{\sqrt{3}}{3}$

Answer 2

Given that it was some sort of quiz, you can guess that:

1. It is on the imaginary axis (because the triangle formed by $1,-1,i$ is isoceles, and see uniqueness of Geometric median on Wikipedia).
2. The point lies inside the isoceles triangle.

Let $f(y)=2\sqrt{1+y^2}+1-y$.

$f'(y)=\frac{2y}{\sqrt{1+y^2}}-1=0\implies4y^2=1+y^2\implies y=\frac{1}{\sqrt3}$.

Answer 3

The post has received good and simple answers based on triangles.

However, if you really want to see the calculus based problem, let me start using Dr. Sonnhard Graubner's approach to the solution. Writing $$f(a,b)=\sqrt{(a+1)^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}$$ Computing the partial derivatives lead to the equations $$\frac{df(a,b)}{da}=\frac{a}{\sqrt{a^2+(b-1)^2}}+\frac{a-1}{\sqrt{(a-1)^2+b^2}}+\frac{a+1}{\sqrt{(a+1)^2+ b^2}}=0$$ $$\frac{df(a,b)}{db}=\frac{b-1}{\sqrt{a^2+(b-1)^2}}+\frac{b}{\sqrt{(a-1)^2+b^2}}+\frac{b}{\sqrt{(a+1)^2+b^ 2}}=0$$ which are extremely complex to solve except if you know (or are able to identify) that one solution corresponds to $a=0$. In such a case, the first partial is effectively equal to $0$ and the second one becomes $$\frac{df(0,b)}{db}=\frac{2 b}{\sqrt{b^2+1}}+\frac{b-1}{\sqrt{(b-1)^2}}=\frac{2 b}{\sqrt{b^2+1}}\pm 1=0$$ for which the solutions are $\mp \frac{1}{\sqrt{3}}$. To these solutions correspond $$f(0,-\frac{1}{\sqrt{3}})=1+\frac{5}{\sqrt{3}}$$ $$f(0,\frac{1}{\sqrt{3}})=1+\sqrt{3}$$ and then the result ($a=0,b=\frac{1}{\sqrt{3}},f=1+\sqrt{3}$).

You can notice that, for $b=\pm \frac{1}{\sqrt{3}}$, $$\frac{d^2f(0,b)}{db^2}=\frac{3 \sqrt{3}}{4} \gt 0$$

Please, notice that the problem with $|z+\alpha|+|z+\beta|+|z-i|$ will be extremely difficult to solve as soon as $\alpha+\beta \neq 0$.

Answer 4

this is called Fermat(?) problem. the minimum occur at a point where every vertex make $120^\circ.$ this is true for any three vertices of a triangle. here we have nicer isosceles triangle of sides $\sqrt 2, \sqrt 2$ and $2.$ the Fermat point is at $\sqrt 3/3.$ the sum of segments is $$2\sqrt 3 /3 + 2\sqrt 3 /3 + 1 - \sqrt 3/3 = 1 + \sqrt 3$$

here is a quick of the proof why Fermat point is where the minimum occurs. Given three non collinear points $A, B, C$ the $$min_{P}\left(|AP| + |BP| + |CP|\right) = |AF|+|BF|+|CF| \mbox{ where } F \mbox{ is such that} \angle AFB = \angle BFC = 120^\circ $$

the proof uses the Ptolemy's theorem that says that in any quadrilateral, the sum of the products of the opposite sides is greater or equal to the product of the diagonals. the equality occurs if and only if the quadrilateral is cyclic.

Proof: Construct exterior equilateral triangles $ABC^\prime, BCA^\prime$ and $CAB^\prime.$ Let the circumcircles of the equilateral triangles meet at the Fermat point $F.$ Let $G$ be any point other than $F.$ Applying Ptolemy's theorem to the cyclic quadrilateral $AFBC^\prime$ and the quadrilateral $AGBC^\prime$ we get $$AF + BF = C^\prime F \mbox{ and } AG + BG \ge C^\prime F$$

adding two similar equations proves the claim.