2021 Fall AMC 12A Problems/Problem 15

Problem

Recall that the conjugate of the complex number $w = a + bi$ , where $a$ and $b$ are real numbers and $i = \sqrt{-1}$ , is the complex number $\overline{w} = a - bi$ . For any complex number $z$ , let $f(z) = 4i\hspace{1pt}\overline{z}$ . The polynomial $P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1$ has four complex roots: $z_1$ , $z_2$ , $z_3$ , and $z_4$ . Let $Q(z) = z^4 + Az^3 + Bz^2 + Cz + D$ be the polynomial whose roots are $f(z_1)$ , $f(z_2)$ , $f(z_3)$ , and $f(z_4)$ , where the coefficients $A,$ $B,$ $C,$ and $D$ are complex numbers. What is $B + D?$

$(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304$

Solution 1

By Vieta's formulas, $z_1z_2z_3z_4=1$ , and $D= (4i)^4\overline{z}_1\,\overline{z}_2\,\overline{z}_3\,\overline{z}_4.$

Since $\overline{a}\cdot\overline{b}=\overline{ab},$ $D=(4i)^4\overline{z_1z_2z_3z_4} = 256(\overline{1}) = 256$

By Vieta's formulas, $z_1z_2+z_1z_3+\dots+z_3z_4=3$ , and $B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).$

Since $\overline{a}\cdot\overline{b}=\overline{ab},$ $B=(4i)^2\left(\overline{z_1z_2}+\overline{z_1z_3}+\overline{z_1z_4}+\overline{z_2z_3}+\overline{z_2z_4}+\overline{z_3z_4}\right).$ Since $\overline{a}+\overline{b}=\overline{a+b},$ $B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48$

Our answer is $B+D=256-48=\boxed{(\textbf{D}) \: 208}.$

~kingofpineapplz ~sl_hc

Solution 2

Since the coefficients of $P$ are real, the roots of $P$ can also be written as $\overline{z_1}, \overline{z_2}, \overline{z_3}, \overline{z_4}$ . With this observation, it's easy to see that the polynomials $P(z)$ and $Q(4i\hspace{1pt}z)$ have the same roots. Hence, there exists some constant $K$ such that \begin{align*} P(z)=K*Q(4i\hspace{1pt}z) \end{align*}

By comparing coefficients, its easy to see that $K=\frac{1}{(4i)^4}$ . Hence $\frac{B*(4i)^2}{(4i)^4}=3$ and $\frac{D}{(4i)^4}=1$ . Hence $B=-48$ , $D=256$ , so $B+D=208$ and our answer is $\boxed{(\textbf{D}) \: 208}$ .

~tsun26, inspired by mAth_SUN

2021 Fall AMC 12A (Problems • Answer Key • Resources)
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2021 Fall AMC 12A Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

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