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

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== Problem ==
 
== Problem ==
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The [[complex number]] <math>z</math> is equal to <math>9+bi</math>, where <math>b</math> is a [[positive]] [[real number]] and <math>i^{2}=-1</math>.  Given that the imaginary parts of <math>z^{2}</math> and <math>z^{3}</math> are the same, what is <math>b</math> equal to?
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>The [[complex number]] <math>z</math> is equal to <math>9+bi</math>, where <math>b</math> is a [[positive]] [[real number]] and <math>i^{2}=-1</math>.  Given that the imaginary parts of <math>z^{2}</math> and <math>z^{3}</math> are the same, what is <math>b</math> equal to?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
 
== Solution ==
 
== Solution ==
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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>.
  
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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>\boxed{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}}
  
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[[Category:Intermediate Complex Numbers Problems]]
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 14:34, 18 February 2017

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 $\boxed{015}$.

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

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