complex numbers
Let $z_1, z_2, z_3, \ldots, z_{13}$ be real numbers, & let $A$ be the average of
complex numbers $[e^{iz_1}, e^{iz_2}, \ldots ,e^{iz_{13}}]$, where
$i=\sqrt{-1}$. As the value of z's vary over all 13-tuples of real numbers,
Find:
i) Maximum value attained by |A|.
ii) Minimum value attained by |A|.
My problem: I know that for a complex no, $x = a+ib$; $|x| = \sqrt{a^2+b^2}$.
But i can not figure out, how to compute | average of the given set of complex no| i.e, |A|.
Please help.
Thank you
Another hint:
Hence, the maximum is $22$ and the minimum is $2$ but you should prove this.
Let $$Q:=(z_0-z_1)^2+(z_1-z_2)^2+(z_2-z_0)^2\in{\mathbb C}\ .$$ The minimal value of $|Q|$ is of course $0$, which is attained when $z_0=z_1=z_2$, but also for an equilateral triangle. In order to determine $\max|Q|$ under the given constraints we may assume $$z_0=e^{it},\quad z_1=-e^{-i\alpha},\quad z_2=-e^{i\alpha}$$ with $t\in{\mathbb R}$ and $0\leq\alpha\leq{\pi\over3}$. Then $$\eqalign{Q&=(e^{it}+e^{i\alpha})^2+(e^{it}+e^{-i\alpha})^2+(2i\sin\alpha)^2 \cr &=2e^{2it}+4e^{it}\cos\alpha+8\cos^2\alpha-6\ . \cr}$$ Put $\cos\alpha=:p\in\bigl[{1\over2},1\bigr]$. Then $$|Q|\leq2+4p+|8p^2-6|\ .$$ If ${\sqrt{3}\over2}\leq p\leq1$ then $$|Q|\leq2+4p+8p^2-6=8\left(p+{1\over4}\right)^2-{9\over2}\leq{25\over2}-{9\over2}=8\ ,$$ and if ${1\over2}\leq p\leq{\sqrt{3}\over2}$ then $$|Q|\leq2+4p+6-8p^2={17\over2}-8\left(p-{1\over4}\right)^2\leq{17\over2}-{1\over2}=8\ .$$ On the other hand $z_0=1$, $z_1=z_2=-1$ gives $|Q|=8$, so that altogether we have proven that $\max|Q|=8$.
Best Answer
Hint: By triangle inequality, then, $$\lvert A\rvert\le\frac1{13}\sum_{k=1}^{13}\left\lvert e^{i z_k}\right\rvert.$$ Since each $z_k$ is real, what can you conclude?
Edit: The kicker, here, is to translate from polar form to rectangular form. In particular, given any real $t,$ we have that $$e^{it}=\cos t+i\sin t,$$ and so Pythagorean Identity lets us explicitly find $\left\lvert e^{it}\right\rvert$ using the definition $\lvert a+ib\rvert=\sqrt{a^2+b^2}$. Can you take it from there?