2021 Fall AMC 12B Problems/Problem 21

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 1

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 root $a_1$ 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 none of the solutions for $a$ have magnitude of $1$ , so the answer is $\boxed{\textbf{(A)}\ 0 }$ ~lopkiloinm

Solution 2

For $\textrm{Im}(P(x))=0$ , we get $\sin(2x)=\sin(x)+\sin(3x)=2\sin(2x)\cos(x)$ So either $\sin(2x)=0$ , i.e. $x\in\{0,\pi\}$ or $\cos(x)=\tfrac 12$ , i.e. $x\in \{\pi/3, 5\pi/3\}$ .

For none of these values do we get $\textrm{Re}(P(x))=0$ .

Therefore, the answer is $\boxed{\textbf{(A) }0}$ .

Solution 3

We have $\begin{align*} P \left( x \right) & = 1 + e^{ix} - e^{i 2x} + e^{i 3x} . \end{align*}$

Denote $y = e^{i x}$ . Hence, this problem asks us to find the number of $y$ with $| y| = 1$ that satisfy $1 + y - y^2 + y^3 = 0 . \hspace{1cm} (1)$

Taking imaginary part of both sides, we have $\begin{align*} 0 & = {\rm Im} \ \left( 1 + y - y^2 + y^3 \right) \\ & = \frac{1}{2i} \left( y - \bar y - y^2 + \bar y^2 + y^3 - \bar y^3 \right) \\ & = \frac{y - \bar y}{2i} \left( 1 - y - \bar y + y^2 + y \bar y + \bar y^2 \right) \\ & = {\rm Im} \ y \left( 1 - \left( y + \bar y \right) + \left( y + \bar y \right)^2 - y \bar y \right) \\ & = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - |y|^2 \right) \\ & = {\rm Im} \ y \left( 1 - 2 {\rm Re} \ y + 4 \left( {\rm Re} \ y \right)^2 - 1 \right) \\ & = 2 {\rm Im} \ y \cdot {\rm Re} \ y \left( 2 {\rm Re} \ y - 1 \right) \\ \end{align*}$ The sixth equality follows from the property that $|y| = 1$ .

Therefore, we have either ${\rm Re} \ y = 0$ or ${\rm Im} \ y = 0$ or $2 {\rm Re} \ y - 1 = 0$ .

Case 1: ${\rm Re} \ y = 0$ .

Because $|y| = 1$ , $y = \pm i$ .

However, these solutions fail to satisfy Equation (1).

Therefore, there is no solution in this case.

Case 2: ${\rm Im} \ y = 0$ .

Because $|y| = 1$ , $y = \pm 1$ .

However, these solutions fail to satisfy Equation (1).

Therefore, there is no solution in this case.

Case 3: $2 {\rm Re} \ y - 1 = 0$ .

Because $|y| = 1$ , $y = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i = e^{i \pm \frac{\pi}{3}}$ .

However, these solutions fail to satisfy Equation (1).

Therefore, there is no solution in this case.

All cases above imply that there is no solution in this problem.

Therefore, the answer is $\boxed{\textbf{(A) }0}$ .

~Steven Chen (www.professorchenedu.com)

Solution 4

Let $a=\cos(x)+i\sin(x)$ , so by De Moivre $P(x)=a^3-a^2+a+1$ . The problem essentially asks for the number of real roots of $P$ which lie on the complex unit circle. Let $|r|=1$ be a root of $P$ , and note that we can't have $r^3-r^2+r=0$ , else $P(r)=0$ . Thus, suppose henceforth that $r^3-r^2+r \neq 0$ . We then have $r^3-r^2+r=r^2(r+\tfrac{1}{r}-1)=a^2(2\mathrm{Re}(r)-1)$ , hence the argument of $r^3-r^2+r$ is either the argument of $a^2$ , or the argument of $-a^2$ . Since $r^3-r^2+r=-1$ is real, it follows that $a^2=\pm 1 \implies a \in \{1,i,-1,-i\}$ . Now, we can check all of these values and find that none of them work, yielding an answer of $\boxed{\textbf{(A) }0}$ .