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2007 AIME I Problems/Problem 3

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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 (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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