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

Revision as of 22:20, 30 November 2007

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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Mock AIME 2 2006-2007

@@ Line 4: / Line 4: @@
 ===Original statement===
-Let <math>\displaystyle n</math> be the smallest positive integer for which there exist positive real numbers <math>\displaystyle a</math> and <math>\displaystyle b</math> such that <math>\displaystyle (a+bi)^n=(a-bi)^n</math>. Compute <math>\displaystyle \frac{b^2}{a^2}</math>.
+Let <math>n</math> be the smallest positive integer for which there exist positive real numbers <math>a</math> and <math>b</math> such that <math>(a+bi)^n=(a-bi)^n</math>. Compute <math>\frac{b^2}{a^2}</math>.
 ==Solution==
@@ Line 18: / Line 18: @@
 *[[Mock AIME 2 2006-2007]]
-[[Category:Intermediate Complex Numbers Problems]]
+[[Category:Intermediate Algebra Problems]]

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