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# Equations and Complex numbers | AIME I, 2019 Question 10

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

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

Equations

Complex Numbers

Integers

AIME, 2019, Question 10

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

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

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

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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