Difference between revisions of "2004 AMC 12B Problems/Problem 16"
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\qquad\mathrm{(E)}\ 8</math> | \qquad\mathrm{(E)}\ 8</math> | ||
| − | == Solution == | + | == Solutions== |
| − | Let <math>z = a+bi</math>, so <math>\overline{z} = a-bi</math>. By definition, <math>z = a+bi = f(z) = i(a-bi) = b+ai</math>, which implies that all solutions to <math>f(z) = z</math> lie on the line <math>y=x</math> on the complex plane. The graph of <math>|z| = 5</math> is a [[circle]] centered at the origin, and there are <math>2 \Rightarrow \mathrm{(C)}</math> intersections. | + | |
| + | ===Solution 1=== | ||
| + | Let <math>z = a+bi</math>, so <math>\overline{z} = a-bi</math>. By definition, <math>z = a+bi = f(z) = i(a-bi) = b+ai</math>, which implies that all solutions to <math>f(z) = z</math> lie on the line <math>y=x</math> on the complex plane. The graph of <math>|z| = 5</math> is a [[circle]] centered at the origin, and there are <math>2 \Rightarrow \mathrm{(C)}</math> intersections. | ||
| + | |||
| + | ===Solution 2=== | ||
| + | We start the same as the above solution: Let <math>z = a+bi</math>, so <math>\overline{z} = a-bi</math>. By definition, <math>z = a+bi = f(z) = i(a-bi) = b+ai</math>. Since we are given <math>|z| = 5</math>, this implies that <math>a^2+b^2=25</math>. We recognize the Pythagorean triple <math>3,4,5</math> so we see that <math>(a,b)=(3,4)</math> or <math>(4,3)</math>. So the answer is <math>2 \Rightarrow \mathrm{(C)}</math>. | ||
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| + | Solution by franzliszt | ||
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| + | ===Solution 3=== | ||
| + | Let <math>z=a+bi</math>, like above. Therefore, <math>z = a+bi = i\overline{z} = i(a-bi) = ai+b</math>. We move some terms around to get <math>bi-b = ai-a</math>. We factor: <math>b(i-1) = a(i-1)</math>. We divide out the common factor to see that <math>b = a</math>. Next we put this into the definition of <math>|z| = a^2 + b^2 = a^2 + a^2 = 2a^2 = 25</math>. Finally, <math>a = \pm\sqrt{\frac{25}{2}}</math>, and <math>a</math> has two solutions. | ||
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 17:57, 16 May 2021
Problem
A function 
is defined by 
, where 
and 
is the complex conjugate of 
. How many values of satisfy both 
and 
?
Solutions
Solution 1
Let 
, so 
. By definition, 
, which implies that all solutions to lie on the line 
on the complex plane. The graph of is a circle centered at the origin, and there are 
intersections.
Solution 2
We start the same as the above solution: Let , so . By definition, . Since we are given , this implies that 
. We recognize the Pythagorean triple 
so we see that 
or 
. So the answer is .
Solution by franzliszt
Solution 3
Let 
, like above. Therefore, 
. We move some terms around to get 
. We factor: 
. We divide out the common factor to see that 
. Next we put this into the definition of 
. Finally, 
, and 
has two solutions.
See also
| 2004 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
