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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 15"
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==Problem 15==
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==Problem==
 
Recall that the conjugate of the complex number <math>w = a + bi</math>, where <math>a</math> and <math>b</math> are real numbers and <math>i = \sqrt{-1}</math>, is the complex number <math>\overline{w} = a - bi</math>. For any complex number <math>z</math>, let <math>f(z) = 4i\hspace{1pt}\overline{z}</math>. The polynomial <cmath>P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1</cmath> has four complex roots: <math>z_1</math>, <math>z_2</math>, <math>z_3</math>, and <math>z_4</math>. Let <cmath>Q(z) = z^4 + Az^3 + Bz^2 + Cz + D</cmath> be the polynomial whose roots are <math>f(z_1)</math>, <math>f(z_2)</math>, <math>f(z_3)</math>, and <math>f(z_4)</math>, where the coefficients <math>A,</math> <math>B,</math> <math>C,</math> and <math>D</math> are complex numbers. What is <math>B + D?</math>
 
Recall that the conjugate of the complex number <math>w = a + bi</math>, where <math>a</math> and <math>b</math> are real numbers and <math>i = \sqrt{-1}</math>, is the complex number <math>\overline{w} = a - bi</math>. For any complex number <math>z</math>, let <math>f(z) = 4i\hspace{1pt}\overline{z}</math>. The polynomial <cmath>P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1</cmath> has four complex roots: <math>z_1</math>, <math>z_2</math>, <math>z_3</math>, and <math>z_4</math>. Let <cmath>Q(z) = z^4 + Az^3 + Bz^2 + Cz + D</cmath> be the polynomial whose roots are <math>f(z_1)</math>, <math>f(z_2)</math>, <math>f(z_3)</math>, and <math>f(z_4)</math>, where the coefficients <math>A,</math> <math>B,</math> <math>C,</math> and <math>D</math> are complex numbers. What is <math>B + D?</math>
  
 
<math>(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304</math>
 
<math>(\textbf{A})\: {-}304\qquad(\textbf{B}) \: {-}208\qquad(\textbf{C}) \: 12i\qquad(\textbf{D}) \: 208\qquad(\textbf{E}) \: 304</math>
  
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==Solution 1==
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==Solution==
  
 
By Vieta's formulas, <math>z_1z_2+z_1z_3+\dots+z_3z_4=3</math>, and <math>B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).</math>
 
By Vieta's formulas, <math>z_1z_2+z_1z_3+\dots+z_3z_4=3</math>, and <math>B=(4i)^2\left(\overline{z}_1\,\overline{z}_2+\overline{z}_1\,\overline{z}_3+\dots+\overline{z}_3\,\overline{z}_4\right).</math>
  
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Since <math>\overline{a}\cdot\overline{b}=\overline{ab},</math>  
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Since <math>\overline{a}\overline{b}=\overline{ab},</math>  
 
 
<cmath>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).</cmath>
 
<cmath>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).</cmath>
 
Since <math>\overline{a}+\overline{b}=\overline{a+b},</math>
 
Since <math>\overline{a}+\overline{b}=\overline{a+b},</math>
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<cmath>B=(4i)^2\overline{\left(z_1z_2+z_1z_3+\dots+z_3z_4\right)}=-16\overline{(3)}=-48</cmath>
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<cmath>B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48</cmath>
  
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Also, <math>z_1z_2z_3z_4=1,</math> and <cmath>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.</cmath>
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Also, <math>z_1z_2z_3z_4=1,</math> and <cmath>D=(4i)^4\left(\overline{z}_1\,\overline{z}_2\,\overline{z}_3\,\overline{z}_4\right)=256\left(\overline{z_1z_2z_3z_4}\right)=256(\overline{1})=256.</cmath>
  
 
Our answer is <math>B+D=256-48=\boxed{(\textbf{D}) \: 208}.</math>
 
Our answer is <math>B+D=256-48=\boxed{(\textbf{D}) \: 208}.</math>
  
 
~kingofpineapplz
 
~kingofpineapplz
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== Solution 2 ==
 
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Because all coefficients of <math>P \left( z \right)</math> are real, <math>\bar z_1</math>, <math>\bar z_2</math>, <math>\bar z_3</math>, and <math>\bar z_4</math> are four zeros of <math>P \left( z \right)</math>.
 
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First, we compute <math>B</math>.
 
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For <math>P \left( z \right)</math>, following from Vieta's formula,
 
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<cmath>
 
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\[
 
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3 = \bar z_1 \bar z_2 + \bar z_1 \bar z_3 + \bar z_1 \bar z_4 + \bar z_2 \bar z_3 + \bar z_2 \bar z_4 + \bar z_3 \bar z_4 .
 
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\]
 
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</cmath>
 
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For <math>Q \left( z \right)</math>, following from Vieta's formula,
 
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<cmath>
 
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\begin{align*}
 
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B & = f \left( z_1 \right) f \left( z_2 \right)
 
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+ f \left( z_1 \right) f \left( z_3 \right)
 
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+ f \left( z_1 \right) f \left( z_4 \right)
 
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+ f \left( z_2 \right) f \left( z_3 \right)
 
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+ f \left( z_2 \right) f \left( z_4 \right)
 
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+ f \left( z_3 \right) f \left( z_4 \right) \\
 
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& = \left( 4 i \right)^2 \left( \bar z_1 \bar z_2 + \bar z_1 \bar z_3 + \bar z_1 \bar z_4 + \bar z_2 \bar z_3 + \bar z_2 \bar z_4 + \bar z_3 \bar z_4 \right) \\
 
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& = - 16 \cdot 3 \\
 
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& = - 48 .
 
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\end{align*}
 
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</cmath>
 
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Second, we compute <math>D</math>.
 
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For <math>P \left( z \right)</math>, following from Vieta's formula,
 
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<cmath>
 
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\[
 
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1 = \bar z_1 \bar z_2 \bar z_3 \bar z_4  .
 
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\]
 
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</cmath>
 
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For <math>Q \left( z \right)</math>, following from Vieta's formula,
 
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<cmath>
 
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\begin{align*}
 
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D & = f \left( z_1 \right) f \left( z_2 \right) f \left( z_3 \right) f \left( z_4 \right) \\
 
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& = \left( 4 i \right)^4 \bar z_1 \bar z_2 \bar z_3 \bar z_4 \\
 
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& = 256 \cdot 1 \\
 
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& = 256 .
 
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\end{align*}
 
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</cmath>
 
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Therefore, <math>B + D = - 48 + 256 = 208</math>.
 
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Therefore, the answer is <math>\boxed{\textbf{(D) }208}</math>.
 
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~Steven Chen (www.professorchenedu.com)
 
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==See Also==
 
==See Also==
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=14|num-a=16}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:01, 24 May 2023

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

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 $\overline{a}+\overline{b}=\overline{a+b},$ \[B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48\]

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

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

~kingofpineapplz

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AMC 12 Problems and Solutions

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