Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 4"

Latest revision as of 09:50, 4 April 2012

Problem

Revised statement

Let $a$ and $b$ be positive real numbers and $n$ a positive integer such that $(a + bi)^n = (a - bi)^n$ , where $n$ is as small as possible and $i = \sqrt{-1}$ . Compute $\frac{b^2}{a^2}$ .

Original statement

Let $n$ be the smallest positive integer for which there exist positive real numbers $a$ and $b$ such that $(a+bi)^n=(a-bi)^n$ . Compute $\frac{b^2}{a^2}$ .

Solution

Two complex numbers are equal if and only if their real parts and imaginary parts are equal. Thus if $(a + bi)^1 = (a - bi)^1$ we have $b = -b$ so $b = 0$ , not a positive number. If $(a + bi)^2 = (a - bi)^2$ we have $2ab = - 2ab$ so $4ab = 0$ so $a = 0$ or $b = 0$ , again violating the givens. $(a + bi)^3 = (a -bi)^3$ is equivalent to $a^3 - 3ab^2 = a^3 - 3ab^2$ and $3a^2b - b^3 = -3a^2b + b^3$ , which are true if and only if $3a^2b = b^3$ so either $b = 0$ or $3a^2 = b^2$ . Thus $n = b^2/a^2 = 3$ .

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+==See Also==
+{{Mock AIME box|year=2006-2007|n=2|num-b=3|num-a=5}}
-*[[Mock AIME 2 2006-2007/Problem 3 | Previous Problem]]
-*[[Mock AIME 2 2006-2007/Problem 5 | Next Problem]]
-*[[Mock AIME 2 2006-2007]]
 [[Category:Intermediate Algebra Problems]]

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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