2021 Fall AMC 12A Problems/Problem 15

Problem 15

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

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

Also, $z_1z_2z_3z_4=1,$ and $D=(4i)^4\left(\overline{z}_1\,\overline{z}_2\,\overline{z}_3\,\overline{z}_4\right)=256\overline{\left(z_1z_2z_3z_4\right)}=256\overline{(1)}=256.$

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

~kingofpineapplz

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2021 Fall AMC 12A Problems/Problem 15

Problem 15

Solution