Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2019 based on Equations and Complex numbers.

Equations and Complex numbers – AIME 2019

For distinct complex numbers \(z_1,z_2,……,z_{673}\) the polynomial \((x-z_1)^{3}(x-z_2)^{3}…..(x-z_{673})^{3}\) can be expressed as \(x^{2019}+20x^{2018}+19x^{2017}+g(x)\), where g(x) is a polynomial with complex coefficients and with degree at most 2016. The value of \(|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|\) can be expressed in the form \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n

  • is 107
  • is 352
  • is 840
  • cannot be determined from the given information

Key Concepts


Complex Numbers


Check the Answer

But try the problem first…

Answer: is 352.

Suggested Reading

AIME, 2019, Question 10

Complex Numbers from A to Z by Titu Andreescue

Try with Hints

First hint

here \(|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|\)=s=\((z_1z_2+z_1z_3+….z_1z_{673})+(z_2z_3+z_2z_4+…+z_2z_{673})\)

\(+…..+(z_{672}z_{673})\) here


Second Hint

with Vieta’s formula,\(z_1+z_1+z_1+z_2+z_2+z_2+…..+z_{673}+z_{673}+z_{673}\)=-20 then \(z_1+z_2+…..+z_{673}=\frac{-20}{3}\) the first equation and \({z_1}^{2}+{z_1}^{2}+{z_1}^{2}+{z_1z_2}+{z_1z_2}+{z_1z_2}+…..\)=\(3({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})\)+\(9({z_1z_2}+{z_1z_3}+….+{z_{672}z_{673}})\)=\(3({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})\)+9s which is second equation

Final Step

here \((z_1+z_2+…..+z_{673})^{2}=\frac{400}{9}\) from second equation then \(({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})+2({z_1z_2}+{z_1z_3}+….+{z_{672}z_{673}})=\frac{400}{9}\) then \(({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})+2s=\frac{400}{9}\) then \(({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})=\frac{400}{9}\)-2s then with second equation and with vieta s formula \(3(\frac{400}{9}-2s)+9s\)=19 then s=\(\frac{-343}{9}\) then |s|=\(\frac{343}{9}\) where 343 and 9 are relatively prime then 343+9=352.


Subscribe to Cheenta at Youtube