Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $|z|\leq 1$.

share|cite|improve this question
@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
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 $|1+z|+|1-z+z^2|=1.7321<\sqrt{\frac72}=1.8708$.

share|cite|improve this answer
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), $\cos^2t\ge\frac34, -\cos t\ge \frac{\sqrt3}2\implies |1+z|+|1-z+z^2|\ge 4\cdot\frac34+2\cdot\frac{\sqrt3}2-3={\sqrt3}$

share|cite|improve this answer
+1. Nice reduction into quadratic cases. – user1551 Dec 23 '12 at 21:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.