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

 
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== Problem ==
 
== Problem ==
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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>.
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===Revised statement===
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Let <math>a</math> and <math>b</math> be [[positive]] [[real number]]s and <math>n</math> a [[positive integer]] such that <math>(a + bi)^n = (a - bi)^n</math>, where <math>n</math> is as small as possible and <math>i = \sqrt{-1}</math>.  Compute <math>\frac{b^2}{a^2}</math>.
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===Original statement===
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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>.
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==Solution==
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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 = b^2/a^2 = 3</math>.
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==See Also==
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{{Mock AIME box|year=2006-2007|n=2|num-b=3|num-a=5}}
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[[Category:Intermediate Algebra Problems]]

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 $n = b^2/a^2 = 3$.


See Also

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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