Difference between revisions of "2004 AMC 12B Problems/Problem 16"
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== Problem == | == Problem == | ||
− | A | + | A function <math>f</math> is defined by <math>f(z) = i\overline{z}</math>, where <math>i=\sqrt{-1}</math> and <math>\overline{z}</math> is the complex conjugate of <math>z</math>. How many values of <math>z</math> satisfy both <math>|z| = 5</math> and <math>f(z) = z</math>? |
<math>\mathrm{(A)}\ 0 | <math>\mathrm{(A)}\ 0 | ||
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===Solution 2=== | ===Solution 2=== | ||
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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. | 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. | ||
Latest revision as of 12:59, 30 March 2023
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
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.