# n complex numbers inside a disk with center $A$ and radius 1

Inside the disk with center $$A(2,0)$$ and a radius of $$1$$, a set of $$n \geq 1$$ points is considered, each having the respective affixes $$z_1, z_2, \ldots, z_n$$. Show that

$$\left| \sum_{i=1}^{n} z_i \right| \cdot \left| \sum_{i=1}^{n} \frac{1}{z_i} \right| > \frac{3}{4} n^2.$$ source: RMG 2023

I have tried writing $$z_k = a_k + b_k \cdot i$$, where $$|z_k - 2|<1$$ and then using the formula for the modulus of a complex number and it got me here

$$\sqrt{\left(\sum_{k=1}^{n} a_k\right)^2 + \left(\sum_{k=1}^{n} b_k\right)^2} \cdot \sqrt{\left(\sum_{k=1}^{n} \frac{a_k}{a_k^2 + b_k^2}\right)^2 + \left(\sum_{k=1}^{n} \frac{b_k}{a_k^2 + b_k^2}\right)^2} > \frac{3}{4}n^2$$ How could I continue from here or what would be a better proof for it?

• What is RMG?please share the full form. Thank you. Oct 6 at 16:22
• @SoumyadipDas Romanian Mathematical Gazette Oct 6 at 16:37
• is it same as the Romanian maths magazine ? Oct 7 at 2:24

First note that $$\tag{1} \sum_{k=1}^n |z_k| \cdot \sum_{k=1}^n \frac{1}{|z_k|} \ge n^2$$ by the inequality between harmonic and arithmetic mean.
What we need next is a lower bound of $$\left| \sum_{k=1}^n z_k \right|$$ in terms of $$\sum_{k=1}^n |z_k|$$, and similarly for the sum of the reciprocals. This is where the restriction $$|z_k - 2|<1$$ comes into play: All $$z_k$$ in that disk lie in the sector $$S = \{ z = r e^{i \phi} \mid r > 0, -\frac\pi 6 < \phi < \frac\pi 6\} \,.$$
For $$z = x+iy \in S$$ is $$|y| \le \tan(\frac \pi 6) \cdot x = \frac{x}{\sqrt 3}$$ and therefore $$|z| = \sqrt{x^2+y^2} \le \sqrt{x^2+ \frac{x^2}{3}} = \sqrt{\frac 4 3} \cdot x \, .$$ It follows that $$\tag{2} \left| \sum_{k=1}^n z_k \right| \ge \operatorname{Re} \left( \sum_{k=1}^n z_k \right) = \sum_{k=1}^n \operatorname{Re}(z_k) \ge \sqrt{\frac 3 4} \sum_{k=1}^n |z_k| \, .$$ Also $$z \in S \implies 1/z \in S$$, so that $$\tag{3} \left| \sum_{k=1}^n \frac{1}{z_k} \right| \ge \sqrt{\frac 3 4} \sum_{k=1}^n \frac{1}{|z_k|}$$ holds as well.
Combining the estimates $$(1)$$, $$(2)$$, and $$(3)$$ we get the desired inequality.
• Could you please explain to me this part: $$S = \{ z = r e^{i \phi} \mid r > 0, -\frac\pi 6 < \phi < \frac\pi 6\} \,.$$ what does $z = r e^{i \phi}$ mean. I have just started working with complex numbers and I’m not very familiar with those notations. Thank you so much in advance Sep 20 at 17:46
• @math.enthusiast9: That are polar coordinates. $e^{i \phi} = r (\cos \phi + i \sin \phi)$, $r$ is the absolute value and $\phi$ the argument of $z$. $S$ is the sector of all complex numbers with an argument between $-\pi/6$ and $\pi/6$. Sep 20 at 17:50