Given $12$ complex numbers $z_1,...,z_{12}$ st for each $1 \leq i \leq 12$: $$|z_i|=2 , |z_i - z_{i+1}| \geq 1$$prove that : $$\sum_{1 \leq i \leq 12} \frac{1}{|z_i\overline{z_{i+1}}+1|^2} \geq \frac{1}{2}$$
Problem
Source: Iran MO 2023 3rd round , Algebra exam P1
Tags: geometry, complex numbers
25.08.2023 21:59
Just note that $|z_i - z_{i+1}|^2 \geq 1$ and $|z_i\overline{z_{i+1}}+1|^2 \leq 24$ are both equivalent to $z_i\overline{z_{i+1}} + z_{i+1}\overline{z_i} \leq 7$ Btw there's another solution , by just considering the points on a circle with diameter $-3 , 5$ after transforming all the points $z_i\overline{z_{i+1}}+1$ on this circle. Here's an sketch: The only points $X$ on that circle which satisfies $OX^2 \leq 24$ , lie on a very small arc with midpoint $5 + 0i$.But by just locating two points $z_i , z_{i+1}$ on $C(0,2)$ st their distance is large enough, on can easily find that the angle $arg(z_i) - arg(z_{i+1})$ is large enough . But $arg(z_i\overline{z_{i+1}})$ is just $arg(z_i) - arg(z_{i+1})$ which was large enough ; that is , larger than the $arg$ of the points of that small arc around $5$.
28.08.2023 01:05
Thanks...
12.09.2023 11:37
There are two cases for zi:negative zi and positive zi Negative zi is -2 Positive zi is 2 Any complex number is possible for the negative case(any zi+1) Any complex number is also possible for the positive case either zi+1=2 . Therefore the inequality is possible .
20.03.2024 01:12
Proposed by Hesam Rajabzadeh.
28.06.2024 23:38
It is well known that $|x|^2=x\cdot \overline{x}$. Then $$1\leq |z_i-z_{i+1}|\Rightarrow 1\leq (z_i-z_{i+1})(\overline{z_i}-\overline{z_{i+1}})\Rightarrow z_i \overline{z_{i+1}}+z_{i+1}\overline{z_i}\leq 7$$Finally $$\frac{1}{|z_i\overline{z_{i+1}}+1|^2}=\frac{1}{(z_i\overline{z_{i+1}}+1)(z_{i+1}\overline{z_{i}}+1)}=\frac{1}{17+z_i \overline{z_{i+1}}+z_{i+1}\overline{z_i}}\geq \frac{1}{17+7}=\frac{1}{24}$$ @ below I think it is sufficient to note that the denominator is strictly positive.
19.08.2024 14:34
sami1618 wrote: It is well known that $|x|^2=x\cdot \overline{x}$. Then $$1\leq |z_i-z_{i+1}|\Rightarrow 1\leq (z_i-z_{i+1})(\overline{z_i}+\overline{z_{i+1}})\Rightarrow z_i \overline{z_{i+1}}+z_{i+1}\overline{z_i}\leq 7$$Finally $$\frac{1}{|z_i\overline{z_{i+1}}+1|^2}=\frac{1}{(z_i\overline{z_{i+1}}+1)(z_{i+1}\overline{z_{i}}+1)}=\frac{1}{17+z_i \overline{z_{i+1}}+z_{i+1}\overline{z_i}}\geq \frac{1}{17+7}=\frac{1}{24}$$ Shouldn't it also be metioned that $ z_i \overline{z_{i+1}}+z_{i+1}\overline{z_i}$ is positve?