Find the minimum value of the expression $|z|^2 +|z-3|^2 +|z-6i|^2$ I have this question as homework: 

Find the minimum value of the expression: $|z|^2 +|z-3|^2 +|z-6i|^2.$

Here's what I did:
I plotted the points (0,0), (0,6), and (3,0) on the argand plane and joined them to make a triangle. 
Now here is where I doubt myself. I found it centroid (1,2) and this should be the centre of mass if unit masses are kept at every vertex. Yes, centre of mass. Now the Moment of Inertia of a planar object is minimum about the axis passing perpendicular through its centre of mass and also from the parallel axis theorem 
$ I = Icom + md^2$
where $ Icom = (1)|z|^2 + (1)|z-3|^2 + (1)|z-6i|^2$
I is the moment of Inertia about any axis parallel to Icom at a distance d from it.
So the minimum value of expression must be minimum about this point only. My answer is also correct.
If I am correct how do to explain it mathematically or using mathematic theorems or axioms. Otherwise how do I do it.
Thanks.
I can shift this to Physics Stack Exchange if this doesn't fit here. Sorry for no LaTeX as I have just started using Stack Exchange from mobile. Please edit whatever required.
 A: We can also do it this way. We write the function in terms of $x$ and $y$, as we know $z=x+iy$. Simplifying the expression, we need to minimise: $$f(x,y) = 2 (x^2+y^2) + (x-3)^2 + (y-6)^2$$
We can then find the critical point(s) by setting $\frac{\partial f}{\partial x} =0$ and $\frac{\partial f} {\partial y} =0$.
A: Your analysis is correct. On the other hand, it admits a rather easy algebraic solution.
This is similar to, but simpler than, Steiner’s problem of finding the point such that the sum of the distances from the vertices of a triangle is minimum.
In your case you want to minimize the sum of the squares of the distances: if you write $z=x+iy$, the function to minimize is
$$
f(x,y)=x^2+y^2+(x-3)^2+y^2+x^2+(y-6)^2
=3x^2+3y^2-6x-12y+27
$$
and we may as well minimize
$$
g(x,y)=x^2-2x+y^2-4y+9=(x-1)^2+(y-2)^2+4
$$
If $z_0$ is the centroid, that is,
$$
z_0=\frac{1}{3}(z_1+z_2+z_3)
$$
you can set $z_i=w_i+z_0$ and the problem becomes to minimize
$$
|w_1+z_0|^2+|w_2+z_0|^2+|w_3+z_0|^2=
|w_1|^2+|w_1|^2+|w_3|^2+3|z_0|^2+\overline{(w_1+w_2+w_3)}z_0+(w_1+w_2+w_3)\overline{z_0}
$$
On the other hand
$$
w_1+w_2+w_3=z_1-z_0+z_2-z_0+z_3-z_0=0
$$
so the problem is to minimize
$$
|w_1|^2+|w_1|^2+|w_3|^2+3|z_0|^2
$$
which is obvious.
A: Let $z=x+yi$, where $\{x,y\}\subset\mathbb R$.
Thus, $$|z|^2+|z-3|^2+|z-6i|^2=$$
$$=x^2+y^2+(x-3)^2+y^2+x^2+(y-6)^2=3(x^2+y^2-2x-4y+15)=$$
$$=3((x-1)^2+(y-2)^2+10)\geq30.$$
The equality occurs for $z=1+2i$, which says that $30$ is a minimal value.
