6
$\begingroup$

Given positive integer $n$. Find all integers can be expressed as $$S=\left (\sum_{k=1}^nz_k\right )\left (\sum_{k=1}^n \frac{1}{z_k}\right )$$ for some complex number $z_1,\cdots,z_n$ and $|z_1|=\cdots|z_n|=1$.


Here is my approach. For $n=1$ it's trivial. For $n\ge 2$, let $z_k=e^{it_k}$ where $0<t_k\le 2\pi$. I have prove that $0\le S\le n^2$ and $$S = \left (\sum_{k=1}^n \cos(t_k)\right )^2 + \left (\sum_{k=1}^n \sin(t_k)\right )^2.$$ I have found that for perfect square numbers. For example, if $n=2k$ for $k$ positive integer:

  • $t_1=t_2=\cdots=t_n=0\implies S=4k^2.$
  • $t_1=t_2=\cdots=t_x=0$ and $t_{x+1}=\cdots=t_{2k}=\pi$ $\implies S=0$.
  • $t_1=t_2=\cdots=t_x=0$ and $t_{x+1}=\cdots=t_{2k}=\pi$ $\implies S=(2k-2x)^2$.
  • $t_1=t_2=\cdots=t_{2x}=0$ and $$\left (t_{2x+1},t_{2x+2},t_{2x+3},t_{2x+4},\cdots,t_{2k-1},t_{2k}\right )=\left (\frac{\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},-\frac{\pi}{2},\cdots, \frac{\pi}{2},-\frac{\pi}{2}\right ),$$ we have $S=(2x-1)^2$.

For the rest number I don't have any clue to make construction.

$\endgroup$
5
  • 2
    $\begingroup$ What does $\displaystyle\left (\sum_{k=1}^n\right )\left (\sum_{k=1}^n \frac{1}{z_k}\right )$ mean? $\endgroup$ Apr 3 at 9:20
  • 1
    $\begingroup$ My mistake, now it has been edited $\endgroup$ Apr 3 at 9:49
  • 2
    $\begingroup$ Note that $\left (\sum_{k=1}^n z_k\right )\left (\sum_{k=1}^n 1/z_k\right ) = n+\sum_{1\le p<q\le n} \left (z_p\overline{z_q} + \overline{z_p}\overline{z_q}\right )=n + 2\sum_{1\le p<q\le n}\cos t_p\cos t_q + \sin t_p\sin t_q$. $\endgroup$ Apr 3 at 11:05
  • $\begingroup$ @AnneBauval $(\sum_{k=1}^n \cos(t_k))^2=\sum_{k=1}^n \cos^2 t_k+2\sum_{1\le p<q\le n} \cos t_p \cos t_q$. Similar for $\sin$. $\endgroup$ Apr 3 at 11:59
  • $\begingroup$ Of course, so what? I mean, what allows you to write $S = \left (\sum_{k=1}^n \cos(t_k)\right )^2 + \left (\sum_{k=1}^n \sin(t_k)\right )^2$? And to claim "I have prove that"? $\endgroup$ Apr 3 at 12:01

2 Answers 2

4
$\begingroup$

If $n$ is even, by taking have of $t_k$s equal to $\pi$ and the other half to $0$, we get that $\left (\sum_{k=1}^n \cos(t_k)\right )^2 + \left (\sum_{k=1}^n \sin(t_k)\right)^2$ equals zero. On the other hand, if all $t_k\equiv 0$, we get $n^2$. Since $S$ depends continuously on $t$s, we can thus get all numbers in $[0,n^2]$.

If $n\ge 3$ is odd, choose $t_1=\frac\pi 3$, $t_2=-\frac\pi 3$, $t_3=\pi$ and then the remaining (even number of) $t_k$s alternating $\pi$ and $0$. Then $S=0$. By the same reason, $S$ also takes all real numbers in $[0,n^2]$, $n^2+1$ of which are integers.

Finally, if $n=1$ the answer is clear (1).

$\endgroup$
2
  • 1
    $\begingroup$ Nice. I only wish to add that for this it is crucial that $S(t_1,\ldots,t_n)$ is always a real value. This not immediately obvious if we look at the first OP equation, but it becomes so with the equivalent $\cos$ and $\sin$ equation. If $S$ also took complex values, we could not apply the intermediate values property. $\endgroup$
    – chi
    Apr 3 at 19:49
  • 2
    $\begingroup$ Great answer! I was so focused on finding a construction whose result was an integer that I forgot about its continuity. $\endgroup$ Apr 3 at 19:57
2
$\begingroup$

Here's an explicit construction for all real $0\leq m\leq n^2$, $n\geq 2$.

If $n=2k + l$, $l\in\{0,1\}$, there exists a triangle with sides $(k,k+l,\sqrt{m})$ provided that $l\leq m \leq n^2$ (and allowing degenerate triangles). Let $\phi$ be the measure of the angle opposite the side of length $\sqrt{m}$ as determined by the law of cosines. Let

$$z_i=\begin{cases} 1 & 1 \leq i \leq k+l\\ -e^{-i\phi} & k+l+1 \leq i \leq n \end{cases}\text{.}$$ Then $\lvert\sum_iz_i\rvert^2 = m$.

In particular, there exists a triangle with sides $(1,1,\sqrt{m}+1)$ provided that $0\leq m \leq 1$. Let $\phi$ be the measure of the angle opposite the side of length $\sqrt{m}+1$. Let $\psi$ be the measure of one of the other angles. Let $n\geq 3$ be odd, and let

$$z_i=\begin{cases} 1 & i=1\\ -e^{-i\phi} & i=2\\ -e^{i\psi} & i =3\\ (-1)^i & i >3 \end{cases}\text{.}$$ Then $\lvert\sum_iz_i\rvert^2 = m$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .