the minimum possible value of absolute $|z|^2 + |z-3|^2+|z-6i|^2$,where z is a complex number The minimum possible value of $|z|^2 +|z-3|^2+|z-6i|^2$,where z is a complex number I do not know how to do it. here  $|z|$ refers to absolute value of $z$. Here $i=\sqrt {-1}$.
 A: Initially it reminded me of Fermat point.
But it is much different & simpler,
$$|z|^2 +|z-3|^2+|z-6i|^2=3x^2+3y^2-6x-12y+45$$
$$=3(x^2-2x)+3(y^2-4y)+45$$
$$=3\{(x-1)^2-1\}+3\{(y-2)^2-2^2\}+45\ge45-3-3\cdot4=30$$
for real $x,(x-1)^2\ge0$ and real $y,(y-2)^2\ge0$
A: (A hint)
Your problem can be geometrically interpreted as follows: Given three points $Z_k$ in the plane, for which point $M$ is the sum of the squared distances $|MZ_k|^2$ minimal? Make a reasonable guess and prove it.
A: Hints: putting $\,z=x+iy\;,\;\;x,y\in\Bbb R\;$ , note that you want the minimal value of the two variable function
$$f(x,y):=x^2+y^2+(x-3)^2+y^2+x^2+(y-6)^2=3x^2+3y^2-6x-12y+45\implies$$
$$\begin{align*}f'_x&=6x-6=0\iff& x=1\\
f'_y&=6y-12=0\iff& y=2\end{align*}$$
Thus the only critical point is $\,(1,2)\,$ . Now calculate the Hessian of $\,f\,$ at this point:
$$\begin{align*}f'_{xx}=6\\
f'_{yy}=6\\
f'_{xy}=f'_{yx}=0\end{align*}\implies H_f(1,2)=\begin{vmatrix}6&0\\0&6\end{vmatrix}=36>0\;\;\text{and}\;\;f'_{xx}>0\,,\,\,\text{so}\ldots$$
BTW, I think the minimal value is $\,30\,$
A: Express the function in terms of $x$ and $y$, where $z=x+i y$.  The function to minimize is then
$$f(x,y) = 2 (x^2+y^2) + (x-3)^2 + (y-6)^2$$
Find the critical point(s) by setting $\partial f/\partial x=0$ and $\partial f/\partial y=0$.
