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Difference between revisions of "2000 AIME II Problems/Problem 9"
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Let <math>z=re^{i\theta}</math>. Notice that we have <math>2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{-i\theta}.</math>  
 
Let <math>z=re^{i\theta}</math>. Notice that we have <math>2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{-i\theta}.</math>  
  
−
<math>r</math> must be <math>1</math> (or else if you take the magnitude would not be the same). Therefore, <math>z=e^{i\frac{\pi}{\theta}}</math> and plugging into the desired expression, we get <math>e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=-1</math>. Therefore, the least integer greater is <math>\boxed{000}.</math>
+
<math>r</math> must be <math>1</math> (or if you take the magnitude would not be the same). Therefore, <math>z=e^{i\frac{\pi}{\theta}}</math> and plugging into the desired expression, we get <math>e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=-1</math>. Therefore, the least integer greater is <math>\boxed{000}.</math>
  
 
~solution by williamgolly
 
~solution by williamgolly
Line 24: Line 24:
  
 
== Solution 3 Intuitive ==
 
== Solution 3 Intuitive ==
−
For this solution, we assume that <math>z^{2000} + 1/z^{2000}</math> and <math>z^{2048} + 1/z^{2048}</math> have the same least integer greater than their solution. we have <math>z + 1/z = 2\cos 3</math>. Since <math>\cos 3 < 1</math>, <math>2\cos 3 < 2</math>. If we square the equation <math>z + 1/z = 2\cos 3</math>, we get <math>z^2 + 2 + 1/(z^2) = 4\cos^2 3</math>, or <math>z^2 + 1/(z^2) = 4\cos^2 3 - 2</math>. <math>4\cos^2 3 - 2</math> is is less than <math>2</math>, since <math>4\cos^2 3</math> is less than <math>4</math>. if we square the equation again, we get <math>z^4 + 1/(z^4) = (4\cos^2 3 - 2)^2 -2</math>. since <math>4\cos^2 3 - 2</math> is less than 2, <math>(4\cos^2 3 - 2)^2</math> is less than 4, and <math>(4\cos^2 3 - 2)^2 -2</math> is less than 2. However <math>(4\cos^2 3 - 2)^2 -2</math> is also less than <math>4\cos^2 3 - 2</math>. we can see that every time we square the equation, the right hand side gets smaller, and into the negatives. Since the smallest integer that is allowed as an answer is 0, thus smallest integer greater is <math>\boxed{000}.</math>
 
  
−
~ PaperMath
+
For this solution, we assume that <math>z^{2000} + 1/z^{2000}</math> and <math>z^{2048} + 1/z^{2048}</math> has the same least integer greater than their solution.
 +
 
 +
We have <math>z + 1/z = 2\cos 3</math>. Since <math>\cos 3<1</math>, <math>2\cos 3<2</math>. If we square the equation <math>z + 1/z = 2\cos 3</math>, we get <math>z^2 + 2 + 1/(z^2) = 4\cos^2 3</math>, or <math>z^2 + 1/(z^2) = 4\cos^2 3 - 2</math>. <math>4\cos^2 3 - 2</math> is is less than <math>2</math>, since <math>4\cos^2 3</math> is less than <math>4</math>. If we square the equation again, we get <math>z^4 + 1/(z^4) = (4\cos^2 3 - 2)^2 -2</math>.
 +
 
 +
Since <math>4\cos^2 3 - 2</math> is less than 2, <math>(4\cos^2 3 - 2)^2</math> is less than 4, and <math>(4\cos^2 3 - 2)^2 -2</math> is less than 2. However <math>(4\cos^2 3 - 2)^2 -2</math> is also less than <math>4\cos^2 3 - 2</math>. we can see that every time we square the equation, the right-hand side gets smaller and into the negatives. Since the smallest integer that is allowed as an answer is 0, the smallest integer greater is <math>\boxed{000}.</math>
 +
 
 +
~ PaperMath ~megaboy6679
  
 
== Solution 4 ==
 
== Solution 4 ==
−
First, let <math>z = a+bi</math> where <math>a</math> and <math>b</math> are real numbers. We now have that <cmath>a+bi + \frac{a-bi}{a^2+b^2} = 2 \cos{3^{\circ}}</cmath> given the coniditons of the problem. Equating imaginary coefficients, we have that <cmath>b \left( 1 - \frac{1}{a^2+b^2}\right) = 0</cmath> giving us that either <math>b=0</math> or <math>|z| = 1</math>. Let's consider the latter case for now.  
+
First, let <math>z = a+bi</math> where <math>a</math> and <math>b</math> are real numbers. We now have that <cmath>a+bi+\frac{a-bi}{a^2+b^2} = 2 \cos{3^{\circ}}</cmath> given the conditions of the problem. Equating imaginary coefficients, we have that <cmath>b \left( 1 - \frac{1}{a^2+b^2}\right) = 0</cmath> giving us that either <math>b=0</math> or <math>|z| = 1</math>. Let's consider the latter case for now.  
  
 
We now know that <math>a^2+b^2=1</math>, so when we equate real coefficients we have that <math>2a = 2 \cos{3^{\circ}}</math>, therefore <math>a = \cos{3^{\circ}}</math>. So, <math>b = \cos{3^{\circ}}</math> and then we can write <math>z = \text{cis}(3)^{\circ}</math>
 
We now know that <math>a^2+b^2=1</math>, so when we equate real coefficients we have that <math>2a = 2 \cos{3^{\circ}}</math>, therefore <math>a = \cos{3^{\circ}}</math>. So, <math>b = \cos{3^{\circ}}</math> and then we can write <math>z = \text{cis}(3)^{\circ}</math>
  
−
By De Moivre's Theorem, <cmath>z^{2000} + \frac{1}{z^{2000}} = \text{cis} (6000)^{\circ} + \text{cis} (-6000)^{\circ}</cmath>. The imaginary parts cancel, leaving us with <math>2 \cos{6000^{\circ}}</math> which is <math>240 \pmod{360}</math>. Therefore, we have it being <math>-1</math> and our answer is <math>\boxed{000}</math>.  
+
By De Moivre's Theorem, <cmath>z^{2000} + \frac{1}{z^{2000}} = \text{cis} (6000)^{\circ} + \text{cis} (-6000)^{\circ}</cmath>. The imaginary parts cancel, leaving us with <math>2 \cos{6000^{\circ}}</math>, which is <math>240 \pmod{360}</math>. Therefore, it is <math>-1</math>, and our answer is <math>\boxed{000}</math>.  
 +
 
 +
Now, if <math>b=0</math> then we have that <math>a+\frac{1}{a} = 2 \cos{3^{\circ}}</math>. Therefore, <math>a</math> is not violating our conditions set above.
 +
 
  
−
Now, if <math>b=0</math> then we have that <math>a+\frac{1}{a} = 2 \cos{3^{\circ}}</math>. Therefore, <math>a</math> is not real violating our conditions set above.
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=II|num-b=8|num-a=10}}
 
{{AIME box|year=2000|n=II|num-b=8|num-a=10}}

Latest revision as of 19:24, 2 November 2024

Problem

Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$, find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$.

Solution

Using the quadratic equation on $z^2 - (2 \cos 3 )z + 1 = 0$, we have $z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}$.

There are other ways we can come to this conclusion. Note that if $z$ is on the unit circle in the complex plane, then $z = e^{i\theta} = \cos \theta + i\sin \theta$ and $\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta$. We have $z+\frac 1z = 2\cos \theta = 2\cos 3^\circ$ and $\theta = 3^\circ$. Alternatively, we could let $z = a + bi$ and solve to get $z=\cos 3^\circ + i\sin 3^\circ$.


Using De Moivre's Theorem we have $z^{2000} = \cos 6000^\circ + i\sin 6000^\circ$, $6000 = 16(360) + 240$, so $z^{2000} = \cos 240^\circ + i\sin 240^\circ$.

We want $z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1$.

Finally, the least integer greater than $-1$ is $\boxed{000}$.

Solution 2

Let $z=re^{i\theta}$. Notice that we have $2\cos(3^{\circ})=e^{i\frac{\pi}{60}}+e^{-i\frac{\pi}{60}}=re^{i\theta}+\frac{1}{r}e^{-i\theta}.$

$r$ must be $1$ (or if you take the magnitude would not be the same). Therefore, $z=e^{i\frac{\pi}{\theta}}$ and plugging into the desired expression, we get $e^{i\frac{100\pi}{3}}+e^{-i\frac{100\pi}{3}}=2\cos{\frac{100\pi}{3}}=-1$. Therefore, the least integer greater is $\boxed{000}.$

~solution by williamgolly


Solution 3 Intuitive

For this solution, we assume that $z^{2000} + 1/z^{2000}$ and $z^{2048} + 1/z^{2048}$ has the same least integer greater than their solution.

We have $z + 1/z = 2\cos 3$. Since $\cos 3<1$, $2\cos 3<2$. If we square the equation $z + 1/z = 2\cos 3$, we get $z^2 + 2 + 1/(z^2) = 4\cos^2 3$, or $z^2 + 1/(z^2) = 4\cos^2 3 - 2$. $4\cos^2 3 - 2$ is is less than $2$, since $4\cos^2 3$ is less than $4$. If we square the equation again, we get $z^4 + 1/(z^4) = (4\cos^2 3 - 2)^2 -2$.

Since $4\cos^2 3 - 2$ is less than 2, $(4\cos^2 3 - 2)^2$ is less than 4, and $(4\cos^2 3 - 2)^2 -2$ is less than 2. However $(4\cos^2 3 - 2)^2 -2$ is also less than $4\cos^2 3 - 2$. we can see that every time we square the equation, the right-hand side gets smaller and into the negatives. Since the smallest integer that is allowed as an answer is 0, the smallest integer greater is $\boxed{000}.$

~ PaperMath ~megaboy6679

Solution 4

First, let $z = a+bi$ where $a$ and $b$ are real numbers. We now have that \[a+bi+\frac{a-bi}{a^2+b^2} = 2 \cos{3^{\circ}}\] given the conditions of the problem. Equating imaginary coefficients, we have that \[b \left( 1 - \frac{1}{a^2+b^2}\right) = 0\] giving us that either $b=0$ or $|z| = 1$. Let's consider the latter case for now.

We now know that $a^2+b^2=1$, so when we equate real coefficients we have that $2a = 2 \cos{3^{\circ}}$, therefore $a = \cos{3^{\circ}}$. So, $b = \cos{3^{\circ}}$ and then we can write $z = \text{cis}(3)^{\circ}$

By De Moivre's Theorem, \[z^{2000} + \frac{1}{z^{2000}} = \text{cis} (6000)^{\circ} + \text{cis} (-6000)^{\circ}\]. The imaginary parts cancel, leaving us with $2 \cos{6000^{\circ}}$, which is $240 \pmod{360}$. Therefore, it is $-1$, and our answer is $\boxed{000}$.

Now, if $b=0$ then we have that $a+\frac{1}{a} = 2 \cos{3^{\circ}}$. Therefore, $a$ is not violating our conditions set above.


See also

2000 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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