Difference between revisions of "2021 Fall AMC 12A Problems/Problem 15"
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<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> | ||
− | ==Solution | + | ==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> | ||
− | + | Since <math>\overline{a}\cdot\overline{b}=\overline{ab},</math> | |
− | 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> |
Revision as of 02:00, 26 November 2021
Contents
Problem 15
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
Solution 2
Because all coefficients of are real,
,
,
, and
are four zeros of
.
First, we compute .
For , following from Vieta's formula,
For , following from Vieta's formula,
Second, we compute .
For , following from Vieta's formula,
For , following from Vieta's formula,
Therefore, .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
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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