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
R-sk
24.11.2020 11:57
Above I think c(n) must be z(n)
Tintarn
24.11.2020 13:56
Write $a_k=z_k$ for $k$ odd and $a_k=iz_k$ for $k$ even so that $a_k \in \mathbb{R}$ all the time. Moreover, the condition now says that $\vert a_1a_2\vert=2$ and $\vert a_{2k+1}\vert=2^k\vert a_1\vert$ as well as $\vert a_{2k}\vert=2^{k-1}\vert a_2\vert$. We now find that
\[f_n^2=(a_1+a_3+\dots+)^2+(a_2+a_4+\dots)^2=a_1^2 \cdot (1 \pm 2 \pm 4 \pm 8 \dots)^2+a_2^2 \cdot (1 \pm 2 \pm 4 \pm \dots)^2.\]First for (1): We can choose the signs arbitrarily on both sides and hence it's easy to see that we can make both alternating sums of powers of $2$ equal to $1$, but not smaller (in absolute value). Hence
\[f_n^2 \ge a_1^2+a_2^2 \ge 2\vert a_1a_2\vert=4\]by AM-GM and hence $f_n \ge 2$ for all $n \ge 2$ with equality achievable for each $n$. So the desired maximum is equal to $2$.
Next for (2): In $f_{2n} \cdot f_{2n+1}$, both terms have the same part for $a_2,a_4,\dots,a_{2n}$. So again here we can choose the signs to minimize both terms which will be achieved at $1$.
For the odd indices we need to be a bit careful and hence write the number achieved from the signs from $a_1,a_3,\dots,a_{2n-1}$ as $x=1 \pm 2 \pm 4 \pm \dots \pm 2^{n-1}$. So $f_{2n}^2 \ge a_1^2+x^2a_2^2$ and $f_{2n+1}^2 \ge a_1^2+(2^n-x)^2a_2^2$.
We see that this becomes certainly minimal only when $x>0$ so that $0<x<2^n$. We then find that
\[f_{2n}^2f_{2n+1}^2 \ge 4(x^2+(2^n-x)^2)+a_1^4+a_1^4+(x(2^n-x))^2a_2^4 \ge 4(x^2+(2^n-x)^2)+8x(2^n-x)=2^{2n+2}\]by AM-GM and the equality can be achieved by choosing $a_1,a_2$ appropriately.
So the minimum value of $f_{2n}f_{2n+1}$ is $2^{n+1}$.
(Similarly, we would find that the minimum value of $f_{2n-1}f_{2n}$ is $2^n$.)
mathaddiction
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.
renrenthehamster
02.12.2020 09:07
If anyone's interested, I did a video solution here: https://www.youtube.com/watch?v=9uRaI9sAJ8Q.
Mathmatin
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
Tintarn
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?)
lifeismathematics
14.12.2022 20:59
Take $z_{1}=r_{1},z_{2}=ir_{2},z_{3}=r_{3},\cdots r_{2n-1}=r_{2n-1} , r_{2n}$ for each $\{r_{2i-1}\}_{i=1}^n$ and $ \{r_{2i}\}_{i=1}^n \in \mathbb{R}$
a) we have $f_{2n}= |r_{1}+r_{3}+\cdots +r_{2n-1}+i(r_{2}+r_{4}+\cdots+r_{2n})|=\sqrt{\left( \sum_{i=1}^n r_{2i-1} \right)^2+\left( \sum_{i=1}^n r_{2i} \right)^2}$
Also we have $|r_{1}\cdot ir_{2}|=2 \implies |r_{1}r_{2}|=2$, similarly $|r_{2}r_{3}|=2^2 , \cdots , |r_{2n-1}r_{2n}|=2^{2n-1}$
so for $n=1010$ we get:
$f_{2020}=\sqrt{r_{1}^2(1\pm 2\pm 4\pm 8+\cdots \pm 2^{1009})^2+r_{2}^2(1\pm 2\pm 4\pm 8+\cdots \pm 2^{1009})^2}$
claim:-for minimal we can adjust $(1\pm 2\pm 4\pm 8\cdots \pm 2^{1009})^2$ to be $1$
pf:- we can make it to be a G.P. for minimal sum, a possible construction is:
$(1+2+4+\cdots +2^{1008}-2^{1009})^2=(2^{1009}-1-2^{1009})^2=1$
Hence $f_{2020} \ge \sqrt{r_{1}^2+r_{2}^2}=\sqrt{r_{1}^2+\frac{4}{r_{1}^2}} \ge 2$ , hence minimum of $f_{2020}$ is $\boxed{2} \qquad \blacksquare$
b) $f_{2020}\cdot f_{2021} =\sqrt{\left(\sum_{i=1}^{1010} r_{2i-1}\right)^2+\left(\sum_{i=1}^{1010} r_{2i}\right)^2}\cdot \sqrt{\left(\sum_{i=1}^{1011} r_{2i-1}\right)^2+\left(\sum_{i=1}^{1010} r_{2i}\right)^2}$
set $\omega=\left|\frac{\sum_{i=1}^{1010} r_{i}}{r_{1}} \right|$ and $\Omega=\left|\frac{\sum_{i=1}^{1011} r_{i}}{r_{1}} \right|$
we get $f_{2020}\cdot f_{2021}=\sqrt{(\omega^2r_{1}^2+r_{2}^2)(\Omega^2r_{1}^2+r_{2}^2)}=\sqrt{\omega^2\Omega^2a_{1}^4+\omega^2a_{1}^2b_{1}^2+b_{1}^2a_{1}^2\Omega^2+b_{1}^4}\ge \sqrt{4(\omega^2+\Omega^2)+8\omega\Omega}=2|\omega+\Omega|$
now we have $|\omega+\Omega| =\frac{1}{|a_{1}|}\cdot \left (|a_{1}+a_{2}+\cdots a_{1010}|+|a_{1}+a_{2}+\cdots a_{1011}| \right)\ge \frac{1}{|a_{1}|}\cdot |a_{1011}|=2^{1010}$
hence we get $f_{2020}\cdot f_{2021} \ge \boxed{2^{1011}}$ as desired $\blacksquare$
mihaig
15.12.2022 14:07
A real beauty