2021 Fall AMC 12A Problems/Problem 15
Contents
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 ![\[P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1\]](http://latex.artofproblemsolving.com/f/8/d/f8d40b9bd58e6aac5422bf2a759eb6b7dd200bf6.png)
has four complex roots: 
, 
, 
, and 
. Let ![\[Q(z) = z^4 + Az^3 + Bz^2 + Cz + D\]](http://latex.artofproblemsolving.com/f/2/b/f2b288e8e9137e92c34dca6a04767376f683b241.png)
be the polynomial whose roots are 
, 
, 
, and 
, where the coefficients 
and 
are complex numbers. What is 
Solution 1
By Vieta's formulas, 
, and 
Since 
![\[B=(4i)^2\left(\overline{z_1z_2}+\overline{z_1z_3}+\overline{z_1z_4}+\overline{z_2z_3}+\overline{z_2z_4}+\overline{z_3z_4}\right).\]](http://latex.artofproblemsolving.com/c/f/b/cfbdd420942bcb64c15b7eacf8cb6c8b5708fb28.png)
Since 
![\[B=(4i)^2\left(\overline{z_1z_2+z_1z_3+\dots+z_3z_4}\right)=-16(\overline{3})=-48\]](http://latex.artofproblemsolving.com/a/0/4/a04970824a655ad5cd05c35884b05e22cf9144da.png)
Our answer is 
~kingofpineapplz
Solution 2
Since the coefficients of 
are real, the roots of can also be written as 
. With this observation, it's easy to see that the polynomials 
and 
have the same roots. Hence, there exists some constant 
such that
\begin{align*}
P(z)=KQ(4i\hspace{1pt}z)
\end{align*}
By comparing coefficients its easy to see that 
. Hence 
and 
. Hence 
, 
. So 
and our answer is 
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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