% Written By Enoch Yu
% Downloaded from archive.enochyu.com

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\title{Problem Set 14}
\author{Enoch Yu}
\date{May 2025}

\begin{document}

\section*{2021 AMC 12B Problem 18}
Let $z$ be a complex number satisfying $12|z|^2=2|z+2|^2+|z^2+1|^2+31.$ What is the value of $z+\frac 6z?$
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$\textbf{(A) }-2 \qquad \textbf{(B) }-1 \qquad \textbf{(C) }\frac12\qquad \textbf{(D) }1 \qquad \textbf{(E) }4$
\begin{solution}
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\textbf{Key Word} Completing Squares
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Let $z=a+bi$. Substitution of $z$ could be done.
\begin{align*}
12(a^2+b^2)&=2\{(a+2)^2+b^2\}+\{(a^2-b^2+1)^2+(2ab)^2\}+31 \\
12a^2+12b^2&=2a^2+8a+8+2b^2+a^4+b^4+1-2a^2b^2+2a^2-2b^2+4a^2b^2+31 \\
a^4+b^4-8a^2-12b^2+2a^2b^2+8a+40&=0 \\
(a^2+b^2)^2-12(a^2+b^2)+4(a^2+2a+1)+36&=0 \\
(a^2+b^2-6)^2+4(a+1)^2&=0
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\therefore\  a^2+b^2-6&=0,\ a=-1,\ b=\sqrt{5}
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z+\frac{6}{z}=\frac{a^2-b^2+2abi+6}{a+bi}&=\frac{2-2\sqrt{5}i}{-1+\sqrt{5}i}=\boxed{\textbf{(A)} -2}
\end{align*}
\end{solution}

\section*{2021 AMC 12B Problem 20}
Let $Q(z)$ and $R(z)$ be the unique polynomials such that\[z^{2021}+1=(z^2+z+1)Q(z)+R(z)\]and the degree of $R$ is less than $2.$ What is $R(z)?$
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$\textbf{(A) }{-}z \qquad \textbf{(B) }{-}1 \qquad \textbf{(C) }2021\qquad \textbf{(D) }z+1 \qquad \textbf{(E) }2z+1$
\begin{solution}
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\textbf{Key Word} $x^3-1=(x-1)(x^2+x+1)$
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\textcolor{red}{GENERAL RULE OF THUMB: When you see $x^2+x+1$, you should immediately think of $x^3-1=(x-1)(x^2+x+1)$.} Let $r_1,r_2,r_3$ be the roots of $z^3-1=0$. WLOG, let $r_1\ne1$.
\begin{align*}
    {r_1}^{2021}+1&=0+R(r_1) \\
    r^2+1&=R(r_1) \\
    R(r_1)&=-r_1\ (\because {r_1}^2+r_1+1=0)\\
    &\Rightarrow\boxed{\textbf{(A) }{-}z}\ 
\end{align*}
\end{solution}

\end{document}
