12
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How to find the maximum and minimum value of $\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|$ (where $|z_1|=|z_2|=|z_3|=1$ are complex numbers.) ?

My try:

$$\begin{align}\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right| &\leq |z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2 \\ &\leq (|z_1|+|z_2|)^2 + (|z_2|+|z_3|)^2 + (|z_3|+|z_1|)^2 \\ &\leq 2^2+2^2+2^2 \leq 12\end{align}$$

However the answer given is $8$. Where am I going wrong and how to do it correctly?

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  • $\begingroup$ Note that these are points on the unit circle, so the max distance two points can have is the length of the diameter. However, you are dealing with 3 points. Try to find out how you would arrange them on the unit circle to maximize distance. $\endgroup$ – David Feb 17 '17 at 20:16
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    $\begingroup$ @DougM I think you found the maximum of $|(z_1-z_2)|^2 + |(z_2-z_3)|^2 + |(z_3-z_1)|^2$ rather than the one given in question. I checked on Wolfram Alpha. The correct answer is 8. $\endgroup$ – user400242 Feb 17 '17 at 20:55
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    $\begingroup$ We do not change the absolute value if we multiply with $\overline z_3^2$. Therefore, $|(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2| = |(\overline z_3z_1-\overline z_3z_2)^2+(\overline z_3z_2-1)^2+(1-\overline z_3z_1)^2|$. With the substitution $u_1 =\overline z_3z_1$ and $u_2= \overline z_3z_2$ this reduces the number of variables to 2 (which hopefully makes the problem easier). $\endgroup$ – Reinhard Meier Feb 17 '17 at 21:01
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    $\begingroup$ @DougM I don't think so. In a certain sense, I have already rotated $z_3$ to $1$. I cannot do the same with $z_1$ or $z_2$, too. $\endgroup$ – Reinhard Meier Feb 17 '17 at 21:09
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    $\begingroup$ To address "where am I going wrong": Your calculation that the expression is $\le 12$ is correct. But there is nothing in it that guarantees that $12$ can be reached. And in fact, it cannot. All you've shown is that $12$ is an upper bound for the maximum, not that it is the maximum itself. $\endgroup$ – Paul Sinclair Feb 17 '17 at 22:56
9
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Let $$Q:=(z_0-z_1)^2+(z_1-z_2)^2+(z_2-z_0)^2\in{\mathbb C}\ .$$ The minimal value of $|Q|$ is of course $0$, which is attained when $z_0=z_1=z_2$, but also for an equilateral triangle. In order to determine $\max|Q|$ under the given constraints we may assume $$z_0=e^{it},\quad z_1=-e^{-i\alpha},\quad z_2=-e^{i\alpha}$$ with $t\in{\mathbb R}$ and $0\leq\alpha\leq{\pi\over3}$. Then $$\eqalign{Q&=(e^{it}+e^{i\alpha})^2+(e^{it}+e^{-i\alpha})^2+(2i\sin\alpha)^2 \cr &=2e^{2it}+4e^{it}\cos\alpha+8\cos^2\alpha-6\ . \cr}$$ Put $\cos\alpha=:p\in\bigl[{1\over2},1\bigr]$. Then $$|Q|\leq2+4p+|8p^2-6|\ .$$ If ${\sqrt{3}\over2}\leq p\leq1$ then $$|Q|\leq2+4p+8p^2-6=8\left(p+{1\over4}\right)^2-{9\over2}\leq{25\over2}-{9\over2}=8\ ,$$ and if ${1\over2}\leq p\leq{\sqrt{3}\over2}$ then $$|Q|\leq2+4p+6-8p^2={17\over2}-8\left(p-{1\over4}\right)^2\leq{17\over2}-{1\over2}=8\ .$$ On the other hand $z_0=1$, $z_1=z_2=-1$ gives $|Q|=8$, so that altogether we have proven that $\max|Q|=8$.

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  • $\begingroup$ Why did you assume $$0\leq\alpha\leq{\pi\over3}$$ ? $\endgroup$ – user400242 Feb 18 '17 at 21:58
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    $\begingroup$ @Mystic The smallest angle of triangle $z_0z_1z_2$ can be no larger than $\pi /3\,$. It can be assumed WLOG that that's the angle at $z_0$, which then translates to $2 \alpha \le 2 \pi /3\,$. $\endgroup$ – dxiv Feb 19 '17 at 1:17
3
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Given $$(z_1-z_2)+(z_2-z_3)+(z_3-z_1)=0$$ and $$\left|(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2\right|=\\ \left|z_1^2-2z_1z_2+z_2^2+z_2^2-2z_2z_3+z_3^2+z_3^2-2z_3z_1+z_1^2\right|=\\ 2\left|z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1\right|=\\ 2\left|z_1(z_1-z_2)+z_2(z_2-z_3)+z_3(z_3-z_1)\right|=\\ 2\left|z_1(z_1-z_2)+z_2(z_2-z_3)+z_3(-(z_1-z_2)-(z_2-z_3))\right|=\\ 2\left|(z_1-z_2)(z_1-z_3)+(z_2-z_3)^2)\right|=...$$ replacing $z_1=1$ $$...=2\left|(1-z_2)(1-z_3)+(z_2-z_3)^2)\right|=2\left|(1-z_2)(1-z_3)+(z_2-1+1-z_3)^2\right|=\\ 2\left|(1-z_2)(1-z_3)+(z_2-1)^2+(1-z_3)^2+2(z_2-1)(1-z_3)\right|=\\ 2\left|(1-z_2)(1-z_3)+(z_2-1)^2+(1-z_3)^2-2(1-z_2)(1-z_3)\right|=\\ 2\left|(1-z_2)^2+(1-z_3)^2-(1-z_2)(1-z_3)\right|=...$$ which is $$...=2\left|\frac{(1-z_2)^3+(1-z_3)^3}{1-z_2+1-z_3}\right|=...$$ using law of sines ... $$...=2\left|\frac{2^3\sin^3{\alpha}+2^3\sin^3{\beta}}{2\sin{\alpha}+2\sin{\beta}}\right|=8\left|\frac{\sin^3{\alpha}+\sin^3{\beta}}{\sin{\alpha}+\sin{\beta}}\right|\leq ...\tag{1}$$ both $\alpha, \beta \in (0,\pi)$ (corner cases can be treated individually), which means $$0<\sin{\alpha}\leq 1,0<\sin{\beta}\leq 1$$ or $$0<\sin^3{\alpha}\leq \sin{\alpha}<1,0<\sin^3{\beta}\leq \sin{\beta}<1$$ thus $$0<\sin^3{\alpha} + \sin^3{\beta} \leq \sin{\alpha} + \sin{\beta}$$ and, continuing (1) $$...\leq 8$$

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