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Find the smallest positive real number $k$ such that, given any finite set $z_1,\cdots, z_n$ of complex numbers, all with strictly positive real and imaginary parts, the following inequality holds: $$|z_1+z_2+\cdots+z_n|\geq \frac{1}{k}(|z_1|+|z_2|+\cdots+|z_n|).$$

Answer- $\sqrt{2}$

My Attempt:

First, we take $n=2$. Let $z_i=r_ie^{i\theta_i}$ for $i=1, 2$. Then $$|z_1+z_2|^2=|r_1e^{i\theta_1}+r_2e^{i\theta_2}|^2= r_1^2+r_2^2+r_1r_2e^{i(\theta_1-\theta_2)}+r_1r_2e^{i(\theta_2-\theta_1)}.$$ Also $|z_1|+|z_2|=r_1+r_2.$ Therefore, the given inequality holds if
$$r_1^2+r_2^2+r_1r_2e^{i(\theta_1-\theta_2)}+r_1r_2e^{i(\theta_2-\theta_1)}\geq \frac{1}{k^2}(r_1+r_2)^2$$ $$\implies (k^2-1)(r_1^2+r_2^2)+r_1r_2(k^2 e^{i(\theta_1-\theta_2)}+k^2e^{i(\theta_2-\theta_1)}-2)\geq 0.$$ which holds if $$k^2(e^{i(\theta_1-\theta_2)}+e^{i(\theta_2-\theta_1)})\geq 2$$

I struck at this point. Please help.

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    $\begingroup$ It is not difficult to show that the inequality holds with $k=\sqrt 2$. But I doubt that this is the smallest value if $n \ge 3$. $\endgroup$
    – Martin R
    Sep 4, 2021 at 10:09
  • $\begingroup$ @MartinR Actually it holds for $k=\sqrt{2}$. $\endgroup$ Sep 4, 2021 at 11:07
  • $\begingroup$ @YiorgosS.Smyrlis: Yes, that is what I said. It is also easy to see that this is the best possible constant in the case of two numbers. My misunderstanding was that I thought that the question asks for the best constant for a fixed number $n$ (which looks more difficult to me if $n \ge 3$). $\endgroup$
    – Martin R
    Sep 4, 2021 at 11:35
  • $\begingroup$ This should also directly follow from minkowski or holder if you rewrite the question in terms of its positive parts. $\endgroup$
    – dezdichado
    Sep 15, 2021 at 14:22

1 Answer 1

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Set $w_k=e^{-i\pi/4}z_k$. Then $w_k=|z_k|e^{i\theta_k}$, where $\theta_k\in (-\pi/4,\pi/4)$, and $\cos\theta_k>1/\sqrt{2}$. Thus $$ \mathrm{Re}\,w_k>\frac{1}{\sqrt{2}}|w_k|. $$ Then $$ \left|\sum_{k=1}^n z_k\right|=\left|\sum_{k=1}^n w_k\right|\ge\sum_{k=1}^n \mathrm{Re} \,w_k =\sum_{k=1}^n |w_k|\cos\theta_k >\frac{1}{\sqrt{2}}\sum_{k=1}^n |w_k|=\frac{1}{\sqrt{2}}\sum_{k=1}^n |z_k|. $$

$k=\sqrt{2}$ is the smallest.

If $z_1=1+i\varepsilon$ and $z_2=\varepsilon+i$, $\varepsilon>0$, then

$$ |z_1+z_2|=(1+\varepsilon)\sqrt{2}, \qquad |z_1|+|z_2|=2\sqrt{1+\varepsilon^2}, \quad \frac{|z_1+z_2|}{|z_1|+|z_2|}=\frac{1}{\sqrt{2}}\cdot\frac{1+\varepsilon}{\sqrt{1+\varepsilon^2}} $$ and $$ \inf_{\varepsilon>0}\frac{|z_1+z_2|}{|z_1|+|z_2|}=\frac{1}{\sqrt{2}}. $$

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    $\begingroup$ I think what you tried to mean is the following $:$ $$\begin{align*} \left |\sum\limits_{k=1}^{n} z_k \right | & = \left |\sum\limits_{k=1}^{n} w_k \right | \\ & \geq \text {Re} \left ( \sum\limits_{k=1}^{n} w_k \right ) \\ & = \sum\limits_{k=1}^{n} \text {Re}\ (w_k) \\ & \gt \frac {1} {\sqrt {2}} \sum\limits_{k=1}^{n} |w_k| \\ & = \frac {1} {\sqrt {2}} \sum\limits_{k=1}^{n} |z_k|. \end{align*}$$ Done! BTW very nice answer.+1 $\endgroup$
    – RKC
    Sep 15, 2021 at 6:24

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