How to find the maximum possible value of $\big| (z_1-z_2)^2+(1-z_1)(1-z_2) \big|$ given that $|z_1|=|z_2|=1$ ?
I cannot think of any easy method to do it. Any suggestion?
P.S: Avoid using multivariable calculus.
How to find the maximum possible value of $\big| (z_1-z_2)^2+(1-z_1)(1-z_2) \big|$ given that $|z_1|=|z_2|=1$?
5
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algebra-precalculus
complex-numbers
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0As hinted in my comment to that [other question](http://math.stackexchange.com/questions/2149243/how-to-find-the-maximum-and-minimum-value-of-leftz-1-z-22-z-2-z-32), it could probably be worked out the hard way, by writing everything in polar form then using some trig (since everything can be expressed in terms of half-angles of the two arguments), but it doesn't look pretty. – 2017-02-18
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2Isn't it $6$ when $z_1 = i$ and $z_2 = -i$ – 2017-02-18
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0@user399601 Correct. I don't know why Wolfram Alpha seems to give a wrong result. (http://www.wolframalpha.com/input/?i=maximum+%7C(z1-z2)%5E2-(1-z1)(1-z2)%7C+given+%7Cz1%7C%3D%7Cz2%7C%3D1) is – 2017-02-18
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0@anonymous I had the sign wrong in my original comment. The inequality to prove has a $+$ not a $-$ between the two terms. As for WA, I believe it's assuming $z_{1,2}$ to be real numbers. – 2017-02-18
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0Taking into account the remark of @user399601, why don't you change the 4 into a 6 in your question ? – 2017-02-18
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0@JeanMarie How do we know if 6 is the correct maximum ? – 2017-02-18
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0@JeanMarie Or keep the $4$ but change the sign from $-$ to $+\,$. – 2017-02-18
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1@dxiv I made that change. – 2017-02-18
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1The reason I think this is duplicate $$(z_1-z_2)+(z_2-z_3)+(z_3-z_1)=0$$ and $$\left|(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2\right|=\\ \left|z_1^2-2z_1z_2+z_2^2+z_2^2-2z_2z_3+z_3^2+z_3^2-2z_3z_1+z_1^2\right|=\\ 2\left|z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1\right|=\\ 2\left|z_1(z_1-z_2)+z_2(z_2-z_3)+z_3(z_3-z_1)\right|=\\ 2\left|z_1(z_1-z_2)+z_2(z_2-z_3)+z_3(-(z_1-z_2)-(z_2-z_3))\right|=\\ 2\left|(z_1-z_2)(z_1-z_3)+(z_2-z_3)^2)\right|$$ and replace $z_1=1$ – 2017-02-20