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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.
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==Solution==
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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>.
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==See Also==
 +
[[Polar form]]
 +
 +
[[Cosine]]
 +
 +
[[Sine]]

Latest revision as of 13:05, 11 September 2011

Solution

We put the number in Polar form: $r(cis{\theta})$. Note that $r(cis{\theta})$ is shorthand for $r(\cos{\theta}+i\sin{\theta})$.Then $(a+bi)^{2002}=r^{2002}(cis(2002\theta))=r(cis(-\theta))$. Since $r^{2002}=r$, either $r=0$ or $r=\pm1$. If $r=0$, then we have $(a,b)=(0,0)$. We can now just focus on $r=1$, since $r=-1$ gives the same solutions.

Since $2002\theta\equiv -\theta\bmod{360}$, $2003\theta=360n$ for some $n$. Each value of $n$ between 1 and 2003 gives a unique solution, so $r=1$ has $2003$ solutions for $\theta$. Thus, there are $2003+1=\boxed{2004}$ solutions; $\boxed{\mathbb E}$.

See Also

Polar form

Cosine

Sine

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