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

Revision as of 19:09, 4 July 2013

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 $015$ .

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2007 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2007 AIME I Problems/Problem 3"

Revision as of 19:09, 4 July 2013

Problem

Solution

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