View source for 2007 AIME I Problems/Problem 3

== Problem ==
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?

== Solution ==
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>.

== See also ==
{{AIME box|year=2007|n=I|num-b=2|num-a=4}}

[[Category:Intermediate Algebra Problems]]
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