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This question already has an answer here:

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.

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marked as duplicate by Jean-Claude Arbaut, José Carlos Santos, Aweygan, Sahiba Arora, user223391 Jan 15 '18 at 18:57

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Do you know multivariable calculus? $\endgroup$ – Archis Welankar Jan 8 '18 at 13:32
  • $\begingroup$ No. I am still in high school. $\endgroup$ – CodeBlooded Jan 8 '18 at 13:34
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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 $\frac{\partial f} {\partial y} =0$.

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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.

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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.

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