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How do I prove that $$\sqrt{\frac{7}{2}}\leq |1+z|+|1-z+z^2|\leq 3\sqrt{\frac{7}{6}}$$ for all complex numbers $|z|=1$? I don't really know how to grapple with it. P.. I am extremely sorry, the condition should actually be $|z|=1$ and it was previously incorrectly stated as $|z|\leq 1$.

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@BenjaLim, why did you delete your answer? –  Richard Nash Dec 23 '12 at 13:20
    
Because it's wrong. –  user38268 Dec 23 '12 at 14:05
    
@BenjaLim, I have changed the problem.It was communicated to me wrong. –  Richard Nash Dec 23 '12 at 14:22

2 Answers 2

up vote 3 down vote accepted

The inequality is wrong. A simple simulation experiment reveals that when $z=-\sqrt{\frac34} + \frac i2$, we have $|z|=1$ and $|1+z|+|1-z+z^2|=3.2497>3\sqrt{\frac76}=3.2404$. When $z=\frac12 + i\sqrt{\frac34}$, we have $|z|=1$ and $|1+z|+|1-z+z^2|=1.7321<\sqrt{\frac72}=1.8708$.

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the problem was communicated to me wrong.Please look at the edited version. –  Richard Nash Dec 23 '12 at 14:23
    
@RichardNash Your new inequality is still wrong. My answer already gives counterexamples to both the upper bound and the lower bound. –  user1551 Dec 23 '12 at 14:38
    
I should be more careful. –  Richard Nash Dec 23 '12 at 15:12

Let $z=\cos 2t+i\sin 2t,$

$|1+z|=|1+\cos 2t+i\sin 2t|=|2\cos^2t+i2\sin t\cos t|=|2\cos t||\cos t+i\sin t|=|2\cos t|$

$|1-z+z^2|=|1-(\cos 2t+i\sin 2t)+(\cos 2t+i\sin 2t)^2$ $=|1-(\cos 2t+i\sin 2t)+(\cos 4t+i\sin 4t)|$ $=|2\cos^22t-\cos 2t+i(2\sin2t\cos2t-\sin2t)|$ $=|(2\cos2t-1)(\cos 2t+i\sin2t)|=|(2\cos2t-1)||(\cos 2t+i\sin2t)|$ $=|(2\cos2t-1)|=|4\cos^2t-3|$

So, $|1+z|+|1-z+z^2|=2|\cos t|+|4\cos^2t-3|$

Now $|\cos t|= \cos t$ if $\cos t\ge 0$ else $-\cos t$

and $|4\cos^2t-3|=-(4\cos^2t-3)$ if $\cos^2t\le\frac34$ ie., if $-\frac{\sqrt3}2\le \cos t \le \frac{\sqrt3}2,$ else $=4\cos^2t-3$

(i)If $\cos t\le-\frac{\sqrt3}2, |1+z|+|1-z+z^2|=-2\cos t+4\cos^2t-3$

(ii)If $-\frac{\sqrt3}2< \cos t <0, |1+z|+|1-z+z^2|=-2\cos t-(4\cos^2t-3)$

(iii)If $0\le \cos t \le \frac{\sqrt3}2, |1+z|+|1-z+z^2|=2\cos t-(4\cos^2t-3)$

(iv)If $\cos t > \frac{\sqrt3}2 , |1+z|+|1-z+z^2|=2\cos t+(4\cos^2t-3)$

In (i), $\cos^2t\ge\frac34, -\cos t\ge \frac{\sqrt3}2\implies |1+z|+|1-z+z^2|\ge 4\cdot\frac34+2\cdot\frac{\sqrt3}2-3={\sqrt3}$

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+1. Nice reduction into quadratic cases. –  user1551 Dec 23 '12 at 21:50

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