Difference between revisions of "2008 AMC 12B Problems/Problem 19"
m (Changed all \text{Re} and \text{Im} to \Re and \Im, respectively. Also added "by the Trivial Inequality.") |
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\Im(f(1)) & = i+i\Im(\alpha)+i\Im(\gamma) \\ | \Im(f(1)) & = i+i\Im(\alpha)+i\Im(\gamma) \\ | ||
− | \Im(f(i)) & = -i+i\Re(\alpha)+i\ | + | \Im(f(i)) & = -i+i\Re(\alpha)+i\Im(\gamma) |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 22:28, 31 March 2017
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:
Let and then we know and Therefore which reaches its minimum when by the Trivial Inequality. Thus, the answer is
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 |
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