Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 4"
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== Problem == | == Problem == | ||
+ | ===Revised statement=== | ||
+ | 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=== | ||
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>\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>. | ||
==Solution== | ==Solution== | ||
+ | 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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{{solution}} | {{solution}} | ||
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*[[Mock AIME 2 2006-2007]] | *[[Mock AIME 2 2006-2007]] | ||
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+ | [[Category:Intermediate Complex Numbers Problems]] |
Revision as of 19:54, 15 September 2006
Problem
Revised statement
Let and be positive real numbers and a positive integer such that , where is as small as possible and . Compute .
Original statement
Let be the smallest positive integer for which there exist positive real numbers and such that . Compute .
Solution
Two complex numbers are equal if and only if their real parts and imaginary parts are equal. Thus if we have so , not a positive number. If we have so so or , again violating the givens. is equivalent to and , which are true if and only if so either or . Thus , where .
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