Difference between revisions of "2008 AMC 12B Problems/Problem 19"
(→Solution: latexed answer) |
m |
||
Line 1: | Line 1: | ||
==Problem 19== | ==Problem 19== | ||
− | A function <math>f</math> is defined by <math>f(z) = (4 + i) z^2 + \alpha z + \gamma</math> for all complex numbers <math>z</math>, where <math>\alpha</math> and <math>\gamma</math> are complex numbers and <math>i^2 = - 1</math>. Suppose that <math>f(1)</math> and <math>f(i)</math> are both real. What is the smallest possible value of <math>| \alpha | + |\gamma |</math> | + | A function <math>f</math> is defined by <math>f(z) = (4 + i) z^2 + \alpha z + \gamma</math> for all complex numbers <math>z</math>, where <math>\alpha</math> and <math>\gamma</math> are complex numbers and <math>i^2 = - 1</math>. Suppose that <math>f(1)</math> and <math>f(i)</math> are both real. What is the smallest possible value of <math>| \alpha | + |\gamma |</math> ? |
<math>\textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad</math> | <math>\textbf{(A)} \; 1 \qquad \textbf{(B)} \; \sqrt {2} \qquad \textbf{(C)} \; 2 \qquad \textbf{(D)} \; 2 \sqrt {2} \qquad \textbf{(E)} \; 4 \qquad</math> | ||
Line 7: | Line 7: | ||
We need only concern ourselves with the imaginary portions of <math>f(1)</math> and <math>f(i)</math> (both of which must be 0). These are: | We need only concern ourselves with the imaginary portions of <math>f(1)</math> and <math>f(i)</math> (both of which must be 0). These are: | ||
− | <math>1) f(1) = i+\ | + | <math>1) f(1) = i+\textrm{Im}(\alpha)+\textrm{Im}(\gamma)</math> |
− | <math>2) f(i) = -i+i\ | + | <math>2) f(i) = -i+i\textrm{Re}(\alpha)+\textrm{Im}(\gamma)</math> |
− | |||
− | |||
+ | Since <math>\textrm{Im}(\gamma)</math> appears in both equations, we let it equal 0 to simplify the equations. This yields two single-variable equations. Equation 1 tells us that the imaginary part of <math>\alpha</math> must be <math>-i</math>, and equation 2 tells us that the real part of <math>\alpha</math> must be <math>i/i = 1</math>. Therefore, <math>\alpha = 1-i</math>. There are no restrictions on <math>\textrm{Re}(\gamma)</math>, so to minimize <math>\gamma</math>'s absolute value, we let <math>\textrm{Re}(\gamma) = 0</math>. | ||
<math>| \alpha | + |\gamma | = |1-i| + |0| = \sqrt{2} \Rightarrow \boxed{B}</math>. | <math>| \alpha | + |\gamma | = |1-i| + |0| = \sqrt{2} \Rightarrow \boxed{B}</math>. |
Revision as of 18:06, 1 June 2011
Problem 19
A function is defined by
for all complex numbers
, where
and
are complex numbers and
. Suppose that
and
are both real. What is the smallest possible value of
?
Solution
We need only concern ourselves with the imaginary portions of and
(both of which must be 0). These are:
Since appears in both equations, we let it equal 0 to simplify the equations. This yields two single-variable equations. Equation 1 tells us that the imaginary part of
must be
, and equation 2 tells us that the real part of
must be
. Therefore,
. There are no restrictions on
, so to minimize
's absolute value, we let
.
.
See Also
2008 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 |