Difference between revisions of "2004 AMC 12B Problems/Problem 16"
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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. | 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. | ||
Revision as of 09:47, 8 July 2020
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
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.