AIME 2001 II · 第 14 题

Solution 1 (Solution 2 more in depth)

If $|z| = 1$, then it is evident that both $|z^{28}| = |z^8| = 1$. Then, $z^{28} - z^8 = 1$. Say $z^{28} = a+bi$ and $z^8 = r+si$. Then, $a-r + (b-s)i = 1$, in which through equivalation of real and imaginary parts, $a=r+1$ and $b=s$, in which we have $z^{28} = (1+r)+si$ and $z^8 = r+si$. Then, because $|z^{28}|=|z^8|$, we have $(1+r)^2+s^2 = r^2+s^2$, in which $r=\frac{-1}{2}$. Then, substituting into the definition of $|z^8|$, we see that $s=\pm \frac{\sqrt{3}}{2}$. Thus, we have two cases.

$\textbf{Case 1:}$ $z^{28}=\frac{1}{2} + \frac{\sqrt{3}}{2}i, z^8=\frac{-1}{2}+\frac{\sqrt{3}}{2}i$

In this case, we rewrite everything in terms of $\text{cis}(x)$. Then, $z^{28} = \text{cis}(\frac{\pi}{3})$, $z^8=\text{cis}(\frac{2\pi}{3})$. Then, because each $z_m = \text{cis}(\theta_m)$, we have that $\text{cis}(28\theta)=\text{cis}(\frac{\pi}{3})$, and $\text{cis}(8\theta)=\text{cis}(\frac{2\pi}{3})$. Thus $28\theta \equiv \frac{\pi}{3} \mod 2\pi$, and $8\theta \equiv \frac{2\pi}{3} \mod 2\pi$.

The second congruence gives us $\theta \equiv \frac{\pi}{12} \mod \frac{\pi}{4}$, or $\theta = \frac{\pi}{12} + \frac{\pi}{4}k$. Substitution into the first modulo gives us $\frac{7\pi}{3} + 7\pi k \equiv \frac{\pi}{3} \mod 2\pi$, in which subtracting $\frac{\pi}{3}$ gives us $2\pi+7\pi k\equiv 0 \mod 2\pi$. This only holds if $k=2m$, and thus we have $\theta = \frac{\pi}{12} + \frac{\pi m}{2}$. The solutions in incremental order to this case are $\frac{\pi}{12},\; \frac{7\pi}{12},\; \frac{13\pi}{12},\; \frac{19\pi}{12}$.

$\textbf{Case 2:}$ $z^{28}=\frac{1}{2} - \frac{\sqrt{3}}{2}i, z^8=\frac{-1}{2}-\frac{\sqrt{3}}{2}i$

Once again, conversion into $\text{cis}(x)$ form and using De Moivre's gives us $28\theta \equiv -\frac{\pi}{3} \mod 2\pi$ and $8\theta \equiv -\frac{2\pi}{3} \mod 2\pi$. Once again, the second equation gives us $\theta = -\frac{\pi}{12} + \frac{\pi}{4}k$. Substitution of $\theta$ into the first again gives us $28x=28(-\frac{\pi}{12} + \frac{\pi k}{4})=-\frac{7\pi}{3}+7\pi k$, in which reducing modulo $2\pi$ and adding $\frac{\pi}{3}$, we obtain $-\frac{2\pi}{3}+7\pi k \equiv 0 \pmod{2\pi}$. Once again, $k=2m$, and so $\theta = -\frac{\pi}{12} + \frac{\pi m}{2}$. The solutions for $\theta$ are $\frac{5\pi}{12},\;\frac{11\pi}{12},\; \frac{17\pi}{12},\; \frac{23\pi}{12}$.

Converting everything into degrees and putting in incremental order, we see that each $\theta_m : 1\le m\le 8 = 15^\circ,\;75^\circ,\;105^\circ,\;165^\circ,\;195^\circ,\;255^\circ,\;285^\circ,\;345^\circ$.

Adding each $\theta_{2k} : 1 \le k \le 4$, we obtain a final answer of $75+165+255+345 = \boxed{840}$.

~Pinotation

Solution 2

$z$ can be written in the form $\text{cis\,}\theta$. Rearranging, we find that $\text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1$

Since the real part of $\text{cis\,}{28}\theta$ is one more than the real part of $\text{cis\,} {8}\theta$ and their imaginary parts are equal, it is clear that either $\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}+\frac {\sqrt{3}}{2}i$, or $\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}- \frac{\sqrt{3}}{2}i$

• Case 1  : $\text{cis\,}{28}\theta = \frac{1}{2}+ \frac{\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac{1}{2}+\frac{\sqrt{3}}{2}i$

Setting up and solving equations, $Z^{28}= \text{cis\,}{60^\circ}$ and $Z^8= \text{cis\,}{120^\circ}$, we see that the solutions common to both equations have arguments $15^\circ , 105^\circ, 195^\circ,$ and $\ 285^\circ$. We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.

• Case 2  : $\text{cis\,}{28}\theta = \frac{1}{2} -\frac {\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta = -\frac {1}{2} -\frac{\sqrt{3}}{2}i$

Again setting up equations ($Z^{28}= \text{cis\,}{300^\circ}$ and $Z^{8} = \text{cis\,}{240^\circ}$) we see that the common solutions have arguments of $75^\circ, 165^\circ, 255^\circ,$ and $345^\circ$

Listing all of these values, we find that $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ is equal to $(75 + 165 + 255 + 345) ^\circ$ which is equal to $\boxed {840}$ degrees. We only want the sum of a certain number of theta, not all of it.