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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 $i=\sqrt {-1}$.

Sepideh Bakhoda
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pagal
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    i dont know how to i – pagal Aug 10 '13 at 10:05
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    i dont know how to do it please help me – pagal Aug 10 '13 at 10:06
  • Welcome to Math.SE! Please, consider updating your question to include what you have tried / where you are getting stuck. You will find that people on this site will be significantly faster to help you if you do that; that way, we know exactly what help you need. – Did Aug 10 '13 at 10:21
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    i dont know how to start it – pagal Aug 10 '13 at 10:26
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    Hmmm... Which related results do you know / were you exposed to during the lectures? – Did Aug 10 '13 at 10:28

4 Answers4

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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 $y,(y-2)^2\ge0$

Community
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lab bhattacharjee
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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 $\,30\,$

DonAntonio
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  • Sorry Don, I know how this works. Not accusing you, just expressing frustration over inconsistent judgment of when to show a hint and when to just produce a full result. – Ron Gordon Aug 10 '13 at 10:33
  • I think the minute one shows this question as a two-variable function one then most of the work is done. There are still some things to explain and polish in the above, but for that the huge, "meaty" hint is already given, imo, once one shows the OP's expression as a two variable thing. You're right, not always is easy to know when/how to stop. – DonAntonio Aug 10 '13 at 10:37
  • BTW, Christian Blatter's answer seems to me the "hintier" one of the three existing now, and a rather nice one BTW – DonAntonio Aug 10 '13 at 10:43
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(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.

Christian Blatter
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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 $\partial f/\partial y=0$.

Ron Gordon
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