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1996 AIME Problems/Problem 11

Revision as of 14:22, 23 August 2017 by Lazysnorlax (talk | contribs) (→‎Problem)

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 $z \neq \cis 0$ (Error compiling LaTeX. Unknown error_msg) 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.

See also

1996 AIME (Problems • Answer Key • Resources)
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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