Difference between revisions of "2008 AMC 12B Problems/Problem 19"
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| − | <math>| \alpha | + |\gamma | = |1-i| + |0| = \sqrt{2}</math> | + | <math>| \alpha | + |\gamma | = |1-i| + |0| = \sqrt{2} \Rightarrow \boxed{B}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2008|ab=B|num-b=18|num-a=20}} | ||
Revision as of 12:43, 30 May 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 | |
