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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

Equations

Complex Numbers

Integers

## Check the Answer

But try the problem first…

Answer: is 352.

Source
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

P=$(x-z_1)(x-z_1)(x-z_1)(x-z_2)(x-z_2)(x-z_2)…(x-z_{673})(x-z_{673})(x-z_{673})$

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.

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