Let $\{ z_n \}_{n \ge 1}$ be a sequence of complex numbers, whose odd terms are real, even terms are purely imaginary, and for every positive integer $k$, $|z_k z_{k+1}|=2^k$. Denote $f_n=|z_1+z_2+\cdots+z_n|,$ for $n=1,2,\cdots$ (1) Find the minimum of $f_{2020}$. (2) Find the minimum of $f_{2020} \cdot f_{2021}$.
Problem
Source: 2021 China Mathematical Olympiad, P1
Tags: inequalities, complex numbers
24.11.2020 11:57
Above I think c(n) must be z(n)
24.11.2020 13:56
27.11.2020 11:26
Let $a_n=z_{2n-1}$, $b_ni=z_{2n}$. Obviously $|a_{n+1}|=2|a_n|$ and $|b_{n+1}|=2|b_n|$. (i) The answer is $2$ achieved when $a_n=\sqrt{2},2\sqrt{2},...,2^{1008}\sqrt{2},-2^{1009}\sqrt{2}$ and $b_n=a_n$, then $|f_{2020}|=2$. We now show that $|f_{2020}|\geq 2$. Indeed, we have $$|a_1+a_2+...+a_{1010}|\geq |a_{1010}|-\sum_{i=1}^{1009}|a_i|=(2^{1010}-2^{1009}-...2-1)|a_1|=|a_1|$$Similarly $$|b_1+b_2+...+b_{1010}|\geq|b_1|$$Therefore $$f_{2020}=\sqrt{\left(\sum_{i=1}^{1010}a_i\right)^2+\left(\sum_{i=1}^{1010}b_i\right)^2}\geq\sqrt{a_1^2+b_1^2}\geq\sqrt{2a_1b_1}=2$$(ii) The answer is $2^{1011}$, achieved when $a_1=(\frac{4}{2^{2010}-1})^\frac{1}{4}$, $b_1=\frac{2}{a_1}$ and $$a_n=2^{n-1}a_1, n\leq 1010, b_n=2^{n-1}b_1,n\leq 1009$$and finally $b_{1010}=-2^{1009}b_1$, $a_{1011}=-2^{1009}a_1$. We now show that $f_{2020}\cdot f_{2021}\geq 2^{1011}$. Let $$k_1=|\sum_{i=1}^{1010}\frac{a_i}{a_1}|$$and $$k_2=|\sum_{i=1}^{1011}\frac{a_i}{a_1}|$$Then \begin{align*} f_{2020}\cdot f_{2021}&=\sqrt{\left(\left(\sum_{i=1}^{1010}a_i\right)^2+\left(\sum_{i=1}^{1010}b_i\right)^2\right)\left(\left(\sum_{i=1}^{1011}a_i\right)^2+\left(\sum_{i=1}^{1010}b_i\right)^2\right)}\\&\geq\sqrt{(k_1^2a_1^2+b_1^2)(k_2^2a_1^2+b_1^2)}\\&=\sqrt{k_1^2k_2^2a_1^4+b_1^4+4(k_1^2+k_2^2)}\\ &\geq\sqrt{4k_1^2+4k_2^2+8k_1k_2}\\ &=2|k_1+k_2| \end{align*}Meanwhile we have $$k_1+k_2=\frac{1}{|a_1|}(|a_1+...+a_{1010}|+|a_1+...+a_{1011}|)\geq \frac{1}{|a_1|}|a_{1011}|\geq 2^{1010}$$This completes the proof.
02.12.2020 09:07
If anyone's interested, I did a video solution here: https://www.youtube.com/watch?v=9uRaI9sAJ8Q.
24.01.2021 12:10
Its not so difficult I think . We can find $z_3$,$z_4$,... based on $z_1$,$z_2$ then we go for the inequality . But I couldnt find a good answer for b . Please help me if you know the answer
24.01.2021 15:59
Mathmatin wrote: Its not so difficult I think . We can find $z_3$,$z_4$,... based on $z_1$,$z_2$ then we go for the inequality . But I couldnt find a good answer for b . Please help me if you know the answer Huh? Did you look at the previous posts? (in particular #3 and #5 which have complete solutions?)
14.12.2022 20:59
15.12.2022 14:07
A real beauty