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,


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


1 Answer 1


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?

  • $\begingroup$ This summation equals 1/13[sigma{e^(zk^2)] (k=1 to 13). Now the summation is always a positive value because of the powers of e cannot be negative. Then how should i proceed? $\endgroup$
    – Number
    May 18, 2015 at 9:12
  • $\begingroup$ Powers of $e$ can certainly be negative. Real powers of $e$ can't be negative, but these aren't real powers of $e$. Do you know any other forms for $e^{it}$ when $t$ is real? Perhaps something of the form $a+ib$? $\endgroup$ May 18, 2015 at 11:27
  • $\begingroup$ No.I'm not enough good with complex numbers. $\endgroup$
    – Number
    May 18, 2015 at 13:05
  • $\begingroup$ That's going to be absolutely essential for this exercise, I'm afraid. You're going to need to know how to write $e^z$ in rectangular form for general complex $z,$ or at least how to do it for $e^{iy}$ for real $y$. Check your text and notes to see what the connection is between polar and rectangular form. $\endgroup$ May 18, 2015 at 17:10
  • $\begingroup$ Ok,i'm trying to find out the things you have mentioned.Thank you. $\endgroup$
    – Number
    May 18, 2015 at 19:15

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