Find minimum value in complex plane? 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?
 A: 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 $z=\tfrac{i}{\sqrt{3}}$.
A: 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
$$|AD|+|BD|+|CD|\ge |PA|+|PB|+|PC|=1+\sqrt{3}$$
A: 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 $z = i/\sqrt3$.
