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For $\theta_1,\theta_2,\dots,\theta_{51}\in R$, let $A(\theta_1,\theta_2,\dots,\theta_{51})$ be the average of the complex numbers $e^{i\theta_1},e^{i\theta_2},\dots,e^{i\theta_{51}}$, where $i=\sqrt{-1}$. Find the minimum and maximum value of $|A|$.

On applying the triangle inequality for complex numbers, I got $|A|\leq1$. Therefore maximum value of $|A|$ is 1. To find the minimum value, I put all the angles equal to $\pi$ for which each $e^{i\theta_n}$ equals -1. On further calculation, I got, $|A|$=1. Therefore the minimum value is also one.

Am I correct?

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Let's be a bit more general: Let $n \in \mathbb{N} \setminus\{0, 1\}$ and $\theta_1,...,\theta_n \in \mathbb{R}$ and let $$A:=\frac{1}{n}\sum_{k=1}^{n} \exp(i \theta_k)$$ You've already shown that $$|A| \leqslant 1$$ with the triangle inequality. We can easily show an example for $|A|=1$ with $\theta_i=0$ for all $i=1,...,n$. It's also clear that $|A| \geqslant 0$ from the properties of the absolute value, so we just need to prove that $|A|=0$ is possible. Let $$\omega_k=\exp\left(\frac{2ki \pi}{n}\right)$$ I claim that $$\sum_{k=1}^{n} \omega_k = 0$$ Can you prove it?

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  • $\begingroup$ Why have you written $\omega_k=\exp(\frac{2ki\pi}{n})$? $\endgroup$ – MrAP Apr 6 at 4:48
  • $\begingroup$ @MrAP Because $\omega_k$ is the $n$-th root of the unity, i.e. $\omega_1^n=...=\omega_n^n=1$. It's quite easy to see that for even $n$ their sum is $0$, and it can be proven that it's $0$ for odd $n$ ass well. $\endgroup$ – Botond Apr 6 at 9:06
  • $\begingroup$ I cannot see how "It's quite easy to see that $\dots$ their sum is $0$". Can you please explain that? I only know that the sum of the $n$ $n$-th roots of unity is equal to $0$. $\endgroup$ – MrAP Apr 15 at 11:28
  • $\begingroup$ @MrAP Aren't my omegas the n-th roots of unity? $\endgroup$ – Botond Apr 15 at 11:45
  • $\begingroup$ Yes. But I do not understand how "It's quite easy to see that for even $n$ their sum is $0$". Also I would like to know why have you considered even and odd $n$ separately. $\endgroup$ – MrAP Apr 15 at 15:03

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