Difference between revisions of "2007 AIME I Problems/Problem 3"

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== Solution ==
 
== Solution ==
Squaring, we find that <math>\displaystyle (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>.
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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>.
  
Setting these two equal, we get that <math>\displaystyle 18bi = 243bi - b^3i</math>, so <math>\displaystyle b(b^2 - 225) = 0</math> and <math>\displaystyle b = -15, 0, 15</math>. Since <math>b > 0</math>, the solution is <math>015</math>.
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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>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2007|n=I|num-b=2|num-a=4}}
 
{{AIME box|year=2007|n=I|num-b=2|num-a=4}}
  
[[Category:Intermediate Complex Numbers Problems]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 22:21, 30 November 2007

Problem

The complex number $z$ is equal to $9+bi$, where $b$ is a positive real number and $i^{2}=-1$. Given that the imaginary parts of $z^{2}$ and $z^{3}$ are the same, what is $b$ equal to?

Solution

Squaring, we find that $(9 + bi)^2 = 81 + 18bi - b^2$. Cubing and ignoring the real parts of the result, we find that $(81 + 18bi - b^2)(9 + bi) = \ldots + (9\cdot 18 + 81)bi - b^3i$.

Setting these two equal, we get that $18bi = 243bi - b^3i$, so $b(b^2 - 225) = 0$ and $b = -15, 0, 15$. Since $b > 0$, the solution is $015$.

See also

2007 AIME I (ProblemsAnswer KeyResources)
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