Difference between revisions of "2021 Fall AMC 12A Problems/Problem 15"
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− | ==Problem | + | ==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> | ||
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<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> | ||
− | <cmath>B=(4i)^2\overline{ | + | <cmath>B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48</cmath> |
− | 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{ | + | 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> |
Latest revision as of 19:01, 24 May 2023
Problem
Recall that the conjugate of the complex number , where
and
are real numbers and
, is the complex number
. For any complex number
, let
. The polynomial
has four complex roots:
,
,
, and
. Let
be the polynomial whose roots are
,
,
, and
, where the coefficients
and
are complex numbers. What is
Solution
By Vieta's formulas, , and
Since
Since
Also, and
Our answer is
~kingofpineapplz
See Also
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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