Monday, January 2, 2006

United We Stand! Topic: Complex Numbers/Trigonometry. Level: AIME.

Problem: (2004 AIME1 - #13) The polynomial $P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17}$ has $34$ complex roots of the form $z_k=r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34$, with $00$. Given that $a_1+a_2+a_3+a_4+a_5=m/n$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.

Solution: Factor the first part as a finite geometric series to get

$P(x) = \left(\frac{x^{18}-1}{x-1}\right)^2-x^{17} = \frac{(x^{19}-1)(x^{17}-1)}{(x-1)^2}$.

So our roots are the $19$th and $17$th Roots of Unity except $x = 1$.

Knowing our complex number facts, we know the $n$th Roots of Unity appear in this form:

$ \cos{\frac{2\pi k}{n}}+i\sin{\frac{2\pi k}{n}} $.

Then, looking back at the problem, we know it's asking the sum of the first five $a$'s, which are actually just the $\frac{k}{n}$'s. So we have $n = 17,19$ so the smallest five should be

$\frac{1}{19}+\frac{1}{17}+\frac{2}{19}+\frac{2}{17}+\frac{3}{19} = \frac{159}{323}$

so the answer is $159+323 = 482$. QED.


Comment: Not too bad once you see the factorizations, but you have to remember those or else! And remember DeMoivre too. That's super handy.


Practice Problem: (2005 AIME2 - #9) For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n=\sin (nt) + i \cos (nt)$ true for all real $t$?


  1. sin t + i cos t = i (cos t - i sin t) = i (cos (-t) + i sin (-t) )
    i^n ( cos (-nt) + i sin (-nt) = i ( cos (-nt) + i sin (-nt) )
    i^n = i
    n congruent to 1 mod 4
    249 numbers :)

  2. Err, slight calculation error. It's actually 250. Don't do that on the actual AIME lol.