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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 21"

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Let <math>a=\cos(x)+i\sin(x)</math>. Now <math>P(a)=1+a-a^2+a^3</math>. <math>P(-1)=-2</math> and <math>P(0)=1</math> so there is a real number <math>a_1</math> between <math>-1</math> and <math>0</math>. The other <math>a</math>'s must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex <math>a</math>'s squared is <math>\frac{1}{a_1}</math> which is greater than <math>1</math>. If <math>x</math> is real number then <math>a</math> must have magnitude of <math>1</math>, but all the solutions for <math>a</math> do not have magnitude of <math>1</math>, so the answer is <math>\boxed{(A) 0}</math> ~lopkiloinm
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Let <math>a=\cos(x)+i\sin(x)</math>. Now <math>P(a)=1+a-a^2+a^3</math>. <math>P(-1)=-2</math> and <math>P(0)=1</math> so there is a real number <math>a_1</math> between <math>-1</math> and <math>0</math>. The other <math>a</math>'s must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex <math>a</math>'s squared is <math>\frac{1}{a_1}</math> which is greater than <math>1</math>. If <math>x</math> is real number then <math>a</math> must have magnitude of <math>1</math>, but all the solutions for <math>a</math> do not have magnitude of <math>1</math>, so the answer is <math>\boxed{\textbf{(A)}\ 0 }</math> ~lopkiloinm

Revision as of 02:08, 24 November 2021

Problem

For real numbers $x$, let \[P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)\] where $i = \sqrt{-1}$. For how many values of $x$ with $0\leq x<2\pi$ does \[P(x)=0?\]

$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution

Let $a=\cos(x)+i\sin(x)$. Now $P(a)=1+a-a^2+a^3$. $P(-1)=-2$ and $P(0)=1$ so there is a real number $a_1$ between $-1$ and $0$. The other $a$'s must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex $a$'s squared is $\frac{1}{a_1}$ which is greater than $1$. If $x$ is real number then $a$ must have magnitude of $1$, but all the solutions for $a$ do not have magnitude of $1$, so the answer is $\boxed{\textbf{(A)}\ 0 }$ ~lopkiloinm

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