Difference between revisions of "2007 AIME I Problems/Problem 3"
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Squaring, we find that <math>(9 + bi)^2 = 81 + 18bi - b^2</math>. Cubing and ignoring the real parts of the result, we find that <math>(81 + 18bi - b^2)(9 + bi) = \ldots + (9\cdot 18 + 81)bi - b^3i</math>. | Squaring, we find that <math>(9 + bi)^2 = 81 + 18bi - b^2</math>. Cubing and ignoring the real parts of the result, we find that <math>(81 + 18bi - b^2)(9 + bi) = \ldots + (9\cdot 18 + 81)bi - b^3i</math>. | ||
| − | Setting these two equal, we get that <math>18bi = 243bi - b^3i</math>, so <math>b(b^2 - 225) = 0</math> and <math>b = -15, 0, 15</math>. Since <math>b > 0</math>, the solution is <math>015</math>. | + | Setting these two equal, we get that <math>18bi = 243bi - b^3i</math>, so <math>b(b^2 - 225) = 0</math> and <math>b = -15, 0, 15</math>. Since <math>b > 0</math>, the solution is <math>\boxed{015}</math>. |
== See also == | == See also == | ||
Latest revision as of 14:34, 18 February 2017
Problem
The complex number 
is equal to 
, where 
is a positive real number and 
. Given that the imaginary parts of 
and 
are the same, what is equal to?
Solution
Squaring, we find that 
. Cubing and ignoring the real parts of the result, we find that 
.
Setting these two equal, we get that 
, so 
and 
. Since 
, the solution is 
.
See also
| 2007 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
