Difference between revisions of "2008 AMC 12B Problems/Problem 19"
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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+\textrm{Im}(\alpha)+\textrm{Im}(\gamma)</math> | + | <math>1) f(1) = i+i\textrm{Im}(\alpha)+i\textrm{Im}(\gamma)</math> |
<math>2) f(i) = -i+i\textrm{Re}(\alpha)+i\textrm{Im}(\gamma)</math> | <math>2) f(i) = -i+i\textrm{Re}(\alpha)+i\textrm{Im}(\gamma)</math> | ||
Revision as of 09:27, 30 July 2015
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 | |
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