Difference between revisions of "AoPS Wiki talk:Problem of the Day/September 11, 2011"
(Created page with "We put the number in Polar form: <math>r(cis{\theta})</math>. Then <math>(a+bi)^{2002}=r^{2002}(cis(2002\theta))=r(cis(-\theta))</math>. Since <math>r^{2002}=r</math>, eith...") |
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− | We put the number in [[Polar form]]: <math>r(cis{\theta})</math>. Then <math>(a+bi)^{2002}=r^{2002}(cis(2002\theta))=r(cis(-\theta))</math>. Since <math>r^{2002}=r</math>, either <math>r=0</math> or <math>r=\pm1</math>. If <math>r=0</math>, then we have <math>(a,b)=(0,0)</math>. We can now just focus on <math>r=1</math>, since <math>r=-1</math> gives the same solutions. | + | ==Solution== |
+ | We put the number in [[Polar form]]: <math>r(cis{\theta})</math>. Note that <math>r(cis{\theta})</math> is shorthand for <math>r(\cos{\theta}+i\sin{\theta})</math>.Then <math>(a+bi)^{2002}=r^{2002}(cis(2002\theta))=r(cis(-\theta))</math>. Since <math>r^{2002}=r</math>, either <math>r=0</math> or <math>r=\pm1</math>. If <math>r=0</math>, then we have <math>(a,b)=(0,0)</math>. We can now just focus on <math>r=1</math>, since <math>r=-1</math> gives the same solutions. | ||
Since <math>2002\theta\equiv -\theta\bmod{360}</math>, <math>2003\theta=360n</math> for some <math>n</math>. Each value of <math>n</math> between 1 and 2003 gives a unique solution, so <math>r=1</math> has <math>2003</math> solutions for <math>\theta</math>. Thus, there are <math>2003+1=\boxed{2004}</math> solutions; <math>\boxed{\mathbb E}</math>. | Since <math>2002\theta\equiv -\theta\bmod{360}</math>, <math>2003\theta=360n</math> for some <math>n</math>. Each value of <math>n</math> between 1 and 2003 gives a unique solution, so <math>r=1</math> has <math>2003</math> solutions for <math>\theta</math>. Thus, there are <math>2003+1=\boxed{2004}</math> solutions; <math>\boxed{\mathbb E}</math>. | ||
+ | ==See Also== | ||
+ | [[Polar form]] | ||
+ | |||
+ | [[Cosine]] | ||
+ | |||
+ | [[Sine]] |
Latest revision as of 13:05, 11 September 2011
Solution
We put the number in Polar form: . Note that is shorthand for .Then . Since , either or . If , then we have . We can now just focus on , since gives the same solutions.
Since , for some . Each value of between 1 and 2003 gives a unique solution, so has solutions for . Thus, there are solutions; .