Difference between revisions of "1984 AIME Problems/Problem 8"
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== Problem == | == Problem == | ||
− | The equation <math>z^6+z^3+1=0</math> has | + | The equation <math>z^6+z^3+1=0</math> has complex roots with argument <math>\theta</math> between <math>90^\circ</math> and <math>180^\circ</math> in the [[complex plane]]. Determine the degree measure of <math>\theta</math>. |
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== Solution 1 == | == Solution 1 == | ||
We shall introduce another factor to make the equation easier to solve. If <math>r</math> is a root of <math>z^6+z^3+1</math>, then <math>0=(r^3-1)(r^6+r^3+1)=r^9-1</math>. The polynomial <math>x^9-1</math> has all of its roots with [[absolute value]] <math>1</math> and argument of the form <math>40m^\circ</math> for integer <math>m</math> (the ninth degree [[roots of unity]]). Now we simply need to find the root within the desired range that satisfies our original equation <math>x^6 + x^3 + 1 = 0</math>. | We shall introduce another factor to make the equation easier to solve. If <math>r</math> is a root of <math>z^6+z^3+1</math>, then <math>0=(r^3-1)(r^6+r^3+1)=r^9-1</math>. The polynomial <math>x^9-1</math> has all of its roots with [[absolute value]] <math>1</math> and argument of the form <math>40m^\circ</math> for integer <math>m</math> (the ninth degree [[roots of unity]]). Now we simply need to find the root within the desired range that satisfies our original equation <math>x^6 + x^3 + 1 = 0</math>. | ||
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== Solution 2 == | == Solution 2 == | ||
− | The substitution <math>y=z^3</math> simplifies the equation to <math>y^2+y+1 = 0</math>. Applying the quadratic formula gives roots <math>y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}</math>, which have arguments of <math>120</math> and <math>240,</math> respectively. | + | The substitution <math>y=z^3</math> simplifies the equation to <math>y^2+y+1 = 0</math>. Applying the quadratic formula gives roots <math>y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}</math>, which have arguments of <math>120</math> and <math>240,</math> respectively. |
+ | We can write them as <math>z^3 = \cos 240^\circ + i\sin 240^\circ</math> and <math>z^3 = \cos 120^\circ + i\sin 120^\circ</math>. | ||
+ | So we can use [[De Moivre's theorem]] (which I would suggest looking at if you never heard of it before) to find the fractional roots of the expressions above! | ||
+ | For <math>\cos 240^\circ + i\sin 240</math> we have <math>(\cos 240^\circ + i\sin 240^\circ)^{1/3}</math> <math>\Rightarrow</math> <math>\cos 80^\circ + i\sin 80^\circ, \cos 200^\circ + i\sin200^\circ,</math> and <math>\cos 320^\circ + i\sin320^\circ.</math> | ||
+ | Similarly for <math>(\cos 120^\circ + i\sin 120^\circ)^{1/3}</math>, we have <math>\cos 40^\circ + i\sin 40^\circ, \cos 160^\circ + i\sin 160^\circ,</math> and <math>\cos 280^\circ + i\sin 280^\circ.</math> | ||
+ | The only argument out of all these roots that fits the description is <math>\theta = \boxed{160}</math> | ||
+ | |||
+ | Note: We can add <math>120</math> to the angles of the previous solutions to get new solutions because De Moivre's formula says that <math>(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta</math> and <math>\frac{360}{3} = 120</math>. ~programmeruser | ||
+ | |||
+ | ~ blueballoon | ||
== See also == | == See also == |
Latest revision as of 19:05, 7 February 2023
Contents
Problem
The equation has complex roots with argument
between
and
in the complex plane. Determine the degree measure of
.
Solution 1
We shall introduce another factor to make the equation easier to solve. If is a root of
, then
. The polynomial
has all of its roots with absolute value
and argument of the form
for integer
(the ninth degree roots of unity). Now we simply need to find the root within the desired range that satisfies our original equation
.
This reduces to either
or
. But
can't be
because if
, then
. (When we multiplied by
at the beginning, we introduced some extraneous solutions, and the solution with
was one of them.) This leaves
.
Solution 2
The substitution simplifies the equation to
. Applying the quadratic formula gives roots
, which have arguments of
and
respectively.
We can write them as
and
.
So we can use De Moivre's theorem (which I would suggest looking at if you never heard of it before) to find the fractional roots of the expressions above!
For
we have
and
Similarly for
, we have
and
The only argument out of all these roots that fits the description is
Note: We can add to the angles of the previous solutions to get new solutions because De Moivre's formula says that
and
. ~programmeruser
~ blueballoon
See also
1984 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |