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Difference between revisions of "2002 AMC 12A Problems/Problem 24"

(New page: == Problem == Find the number of ordered pairs of real numbers <math>(a,b)</math> such that <math>(a+bi)^{2002} = a-bi</math>. <math> \text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{...)
 
m (Corrected triviality 2003th --> 2003rd)
Line 21: Line 21:
 
For <math>s=0</math> we get a single solution <math>(a,b)=(0,0)</math>.
 
For <math>s=0</math> we get a single solution <math>(a,b)=(0,0)</math>.
  
−
Let's now assume that <math>s=1</math>. Multiply both sides by <math>a+bi</math>. The left hand side becomes <math>(a+bi)^{2003}</math>, the right hand side becomes <math>(a-bi)(a+bi)=a^2 + b^2 = 1</math>. Hence the solutions for this case are precisely all the <math>2003</math>-th complex roots of unity, and there are <math>2003</math> of those.
+
Let's now assume that <math>s=1</math>. Multiply both sides by <math>a+bi</math>. The left hand side becomes <math>(a+bi)^{2003}</math>, the right hand side becomes <math>(a-bi)(a+bi)=a^2 + b^2 = 1</math>. Hence the solutions for this case are precisely all the <math>2003</math>rd complex roots of unity, and there are <math>2003</math> of those.
  
 
The total number of solutions is therefore <math>1+2003 = \boxed{2004}</math>.
 
The total number of solutions is therefore <math>1+2003 = \boxed{2004}</math>.

Revision as of 02:19, 27 June 2010

Problem

Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$.

$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

Solution

Let $s=\sqrt{a^2+b^2}$ be the magnitude of $a+bi$. Then the magnitude of $(a+bi)^{2002}$ is $s^{2002}$, while the magnitude of $a-bi$ is $s$. We get that $s^{2002}=s$, hence either $s=0$ or $s=1$.

For $s=0$ we get a single solution $(a,b)=(0,0)$.

Let's now assume that $s=1$. Multiply both sides by $a+bi$. The left hand side becomes $(a+bi)^{2003}$, the right hand side becomes $(a-bi)(a+bi)=a^2 + b^2 = 1$. Hence the solutions for this case are precisely all the $2003$rd complex roots of unity, and there are $2003$ of those.

The total number of solutions is therefore $1+2003 = \boxed{2004}$.

See Also

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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