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

Revision as of 09:45, 16 September 2006

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

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 $n = 3$ , where $\frac {b^2}{a^2} = 3$ .

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Mock AIME 2 2006-2007

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 Two [[complex number]]s are equal if and only if their [[real part]]s and [[imaginary part]]s are equal.  Thus if <math>(a + bi)^1 = (a - bi)^1</math> we have <math>b = -b</math> so <math>b = 0</math>, not a positive number.  If <math>(a + bi)^2 = (a - bi)^2</math> we have <math>2ab = - 2ab</math> so <math>4ab = 0</math> so <math>a = 0</math> or <math>b = 0</math>, again violating the givens.  <math>(a + bi)^3 = (a -bi)^3</math> is equivalent to <math>a^3 - 3ab^2 = a^3 - 3ab^2</math> and <math>3a^2b - b^3 = -3a^2b + b^3</math>, which are true if and only if <math>3a^2b = b^3</math> so either <math>b = 0</math> or <math>3a^2 = b^2</math>.  Thus <math>n = 3</math>, where <math>\frac {b^2}{a^2} = 3</math>.
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