Find the maximum and minimum value of $|A|$

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?

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?
• Why have you written $\omega_k=\exp(\frac{2ki\pi}{n})$? – MrAP Apr 6 at 4:48
• @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. – Botond Apr 6 at 9:06
• 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 $0$. – MrAP Apr 15 at 11:28
• 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. – MrAP Apr 15 at 15:03