# Find the maximum sum of real part roots

Let $z_1,z_2,z_3,\dots,z_{12}$ be the 12 zeroes of the polynomial $z^{12}-2^{36}$. For each $j$, let $w_j$ be one of $z_j$ or $i z_j$. Then the maximum possible value of the real part of $\displaystyle\sum_{j=1}^{12} w_j$ can be written as $m+\sqrt{n}$ where $m$ and $n$ are positive integers. Find $m+n$.

ONLY HINTS, NO COMPLETE SOLUTIONS PLEASE, Test Prep

Without factoring, I got:

$z^{12} = 2^{36} = (2e^{2\pi ik})^{36}$

$z = 8e^{6\pi k i}$

I know there has to be something wrong, because this gives $z = 8$ always, which is not correct.

Please explain what I am doing wrong?

• Is the last sentence of the highlighted section "Find $m+n$" supposed to be "Find $m + \sqrt{n}$"? – hardmath Aug 11 '15 at 19:55

You need to start with $z^{12}=2^{36}e^{2ik\pi}$
• Why just $e^{2i k \pi}$? – smeo Aug 11 '15 at 19:49
• Because $2^{36}$ lies on the positive real axis – David Quinn Aug 11 '15 at 19:51
• @DavidQuinn, but even with the latter, $e^{72 \pi i} = 1$ as well. (+1) – Amad27 Aug 11 '15 at 20:00