Find minimum value in complex plane?

Question

Minimum value of $$|z+1|+|z-1|+|z-i|$$ for all $z$ from complex numbers? Is there any particular way to handle these type of problems?

Answer 1

There may be more efficient ways, but here's an approach that doesn't require knowledge of the Fermat point in the comment.

Substituting $z=a+bi$ and using $|x+yi| = \sqrt{x^2+y^2}$ turns this into a two-variable optimization problem in (the real numbers) $a$ and $b$. Rather nasty, however.

Since $|z-c|$ represents the distance from $z$ to $c$, your problem can be interpreted as minimizing the sum of the distances of $z$ to respectively the (complex) numbers $-1$, $1$ and $i$; make a simple sketch. For reasons of symmetry, it is clear that this minimum will lie on the (positive) imaginary axis, meaning we can take $a=0$ in $z=a+bi$ above, reducing the problem to a one-variable optimization problem in the (real) variable $b$ (with $0<b<1$). $$f(z) \overset{z=a+bi}{\underset{a=0}\longrightarrow} f(b) = |bi+1|+|bi-1|+|bi-i|=2\sqrt{b^2+1}-b+1$$ Finding the minimum of $$f(b) = 2\sqrt{b^2+1}-b+1$$ can be done with the usual technique (look for the zero of $f'(b)$) and yields $b=1/\sqrt{3}$ meaning that the initial expression attains its minimal value at $z=\tfrac{i}{\sqrt{3}}$.

Answer 2

Let $z=a+bi$ then $$|z+1|+|z-i|+|z-1|=\sqrt{(a+1)^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}$$ Let $A(-1,0),B(1,0),C(0,1),D(a,b)$,$|AC|=|CB|$,then the Fermat point $P((0,\dfrac{\sqrt{3}}{3})$ so $$|AD|+|BD|+|CD|\ge |PA|+|PB|+|PC|=1+\sqrt{3}$$

Answer 3

If you consider $z$ as a point in the complex plane, the question is asking to minimize the sum of the distances from $z$ to each of the three points $-1+0i$, $1+0i$, and $0+1i$.

That's equivalent to finding a point in the Cartesian plane that minimizes the sum of distances to the vertices of a triangle. The solution to that problem is the first Fermat point of the triangle. The page at MathWorld gives some properties and a construction.

One useful property is that any two vertices of the triangle subtend an angle of $120$ degrees at the Fermat point. It follows that the three points $z$, $0$, $1$ form a $30$-degree right triangle, that the shorter leg of this triangle has length $1/\sqrt3$, and that the answer is $z = i/\sqrt3$.