1996 AIME Problems/Problem 11

Problem

Let $\mathrm {P}$ be the product of the roots of $z^6+z^4+z^3+z^2+1=0$ that have a positive imaginary part, and suppose that $\mathrm {P}=r(\cos{\theta^{\circ}}+i\sin{\theta^{\circ}})$, where $0<r$ and $0\leq \theta <360$. Find $\theta$.

Solution 1

\begin{eqnarray*} 0 &=& z^6 - z + z^4 + z^3 + z^2 + z + 1 = z(z^5 - 1) + \frac{z^5-1}{z-1}\\ 0 &=& \frac{(z^5 - 1)(z(z-1)+1)}{z-1} = \frac{(z^2-z+1)(z^5-1)}{z-1} \end{eqnarray*}


Thus $z^5 = 1, z \neq 1 \Longrightarrow z = \mathrm{cis}\ 72, 144, 216, 288$,

or $z^2 - z + 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{-3}}{2} = \mathrm{cis}\ 60, 300$

(see cis).

Discarding the roots with negative imaginary parts (leaving us with $\mathrm{cis} \theta,\ 0 < \theta < 180$), we are left with $\mathrm{cis}\ 60, 72, 144$; their product is $P = \mathrm{cis} (60 + 72 + 144) = \mathrm{cis} \boxed{276}$.

Solution 2

Let $w =$ the fifth roots of unity, except for $1$. Then $w^6 + w^4 + w^3 + w^2 + 1 = w^4 + w^3 + w^2 + w + 1 = 0$, and since both sides have the fifth roots of unity as roots, we have $z^4 + z^3 + z^2 + z + 1  | z^6 + z^4 + z^3 + z^2 + 1$. Long division quickly gives the other factor to be $z^2 - z + 1$. The solution follows as above.

Solution 3

Divide through by $z^3$. We get the equation $z^3 + \frac {1}{z^3} + z + \frac {1}{z} + 1 = 0$. Let $x = z + \frac {1}{z}$. Then $z^3 + \frac {1}{z^3} = x^3 - 3x$. Our equation is then $x^3 - 3x + x + 1 = x^3 - 2x + 1 = (x - 1)(x^2 + x - 1) = 0$, with solutions $x = 1, \frac { - 1\pm\sqrt {5}}{2}$. For $x = 1$, we get $z = \text{cis}60,\text{cis}300$. For $x = \frac { - 1 + \sqrt {5}}{2}$, we get $z = \text{cis}{72},\text{cis}{292}$ (using exponential form of $\cos$). For $x = \frac { - 1 - \sqrt {5}}{2}$, we get $z = \text{cis}144,\text{cis}216$. The ones with positive imaginary parts are ones where $0\le\theta\le180$, so we have $60 + 72 + 144 = \boxed{276}$.

Solution 4

This is just a slight variation of Solution 1.

We start off by adding $z^5$ to both sides, to get a neat geometric sequence with $a = 1$ and $r = z$, which gives us $\frac{z^7 - 1}{z - 1} = z^5$. From here, multiply by $z - 1$ to both sides, noting that then $z \neq \cos 0 + i\sin 0$ since, then we are multiplying by $0$ which makes it undefined. We now note that $z^7 - 1 = z^6 - z^5 \implies z^7 - z^6 + z^5 = 1$. (This is the part that it becomes almost identical to Solution 1). Factor $z^5$ from the LHS, to get $z^5( z^2 - z + 1) = 1$. Call the set of roots from $z^5$ as $A$, and set of roots from $z^2 - z + 1$ as $B$. We are to find $|A \cup B| = A + B - |A \cap B|$. (Essentially, we are to find the roots that are not common to both equations/sets, or else we are overcounting a root two times, rather than once. Try out some equation to see where this might apply) Thankfully in this case, none of the roots are overcounted. From here, proceed with Solution 1.

Solution 5

We recognize that $z^6+z^5+z^4+z^3+z^2+z+1=\frac{z^7-1}{z-1}$ and strongly wish for the equation to turn into so. Thankfully, we can do this by simultaneously adding and subtracting $z^5$ and $z$ as shown below. $z^6+z^5+z^4+z^3+z^2+z+1-(z^5+z)=0\implies \frac{z^7-1}{z-1}-(z^5+z)=0$

Now, knowing that $z=1$ is not a root, we multiply by $z-1$ to obtain $z^7-1-(z-1)(z^5+z)=0\implies z^7-1-(z^6+z^2-z^5-z)=0\implies z^7-z^6+z^5-z^2+z-1=0$ Now, we see the $z^2+z-1$ and it reminds us of the sum of two cubes. Cleverly factoring, we obtain that.. $z^5(z^2-z)+z^5-(z^2-z+1)=0\implies z^5(z^2-z+1)-(z^2-z+1)=0\implies (z^5-1)(z^2-z+1)=0$.

Now, it is clear that we have two cases to consider.

Case $1$: $z^5-1=0$ We obtain that $z^5=1$ or $z^5=e^{2\pi{n}{i}}$ Obviously, the answers to this case are $e^{ia}, a\in{\frac{2\pi}{5}, \frac{4\pi}{5}}$

Case $2$: $z^2-z+1=0$ Completing the square and then algebra allows us to find that $z=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}$ which has $\arg$ $\frac{\pi}{3}$

Hence, the answer is $\frac{18\pi}{15}+\frac{5\pi}{15}=\frac{23\pi}{15}\cdot\frac{180}{\pi}=\boxed{276}$

Solution 6

Add 1 to both sides of the equation to get $x^6+x^4+x^3+x^2+1+1=1$. We can rearrange to find that $(x^6+x^3+1)+(x^4+x^2+1)=1$. Then, using sum of a geometric series, $\frac{x^9-1}{x^3-1}+\frac{x^6-1}{x^2-1}=1$.

Combining the two terms of the LHS, we get that $\frac{x^{11}-x^9-x^2+1+x^9-x^6-x^3+1}{x^5-x^3-x^2+1}=1$, so $x^{11}-x^6-x^3-x^2+2=x^5-x^3-x^2+1$, and simplifying, we see that $x^{11}-x^6-x^5+1=0$, so by SFFT, $(x^6-1)(x^5-1)=0$. Then, the roots of our polynomial are the fifth and sixth roots of unity. However, looking back at our expression when it had fractions, we realize that if $x$ is a second or third root of unity, it would cause a denominator to be zero, so the roots of our polynomial are the fifth and sixth roots of unity that are not the second or third roots of unity. The only roots in this category are $\mathrm{cis} (60, 72, 144)$, so our desired sum is $\boxed{276}$, and we are done.

-coolak

See also

1996 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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