Dissertação 1 - ITA - 2ª Fase Dia 1 - ITA 2025

Note que ${\mathrm{x}}^2-\mathrm{x}+1=0\to \mathrm{x}=\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i}\to \mathrm{x}=\mathrm{cis}\left(\frac{\mathrm{\pi }}{3}\right) \mathrm{ou} \mathrm{x}=\mathrm{cis}\left(-\frac{\mathrm{\pi }}{3}\right).$

Considere o algoritmo da divisão:

 ${\mathrm{x}}^{57}+{\mathrm{ax}}^{14}+{\mathrm{bx}}^7+1\equiv ({\mathrm{x}}^2-\mathrm{x}+1)\mathrm{q}\left(\mathrm{x}\right)+(2\mathrm{x}+1)$

Para $\mathrm{x}=\mathrm{cis}\left(\frac{\mathrm{\pi }}{3}\right):$

$\mathrm{cis}\left(\frac{57\mathrm{\pi }}{3}\right)+\mathrm{a}·\mathrm{cis}\left(\frac{14\mathrm{\pi }}{3}\right)+\mathrm{b}·\mathrm{cis}\left(\frac{7\mathrm{\pi }}{3}\right)+1\equiv 0·\mathrm{q}\left(\mathrm{x}\right)+\left(2\mathrm{cis}\left(\frac{\mathrm{\pi }}{3}\right)+1\right)$

Simplificando: 

$\mathrm{cis}\left(\mathrm{\pi }\right)+\mathrm{a}·\mathrm{cis}\left(\frac{2\mathrm{\pi }}{3}\right)+\mathrm{b}·\mathrm{cis}\left(\frac{\mathrm{\pi }}{3}\right)+1\equiv 2\mathrm{cis}\left(\frac{\mathrm{\pi }}{3}\right)+1$

Desse modo:

$-1+\mathrm{a}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}\right)+\mathrm{b}\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}\right)+1=2\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i}\right)+1$

Assim: 

$\left(-1-\frac{\mathrm{a}}{2}+\frac{\mathrm{b}}{2}+1\right)+\left(\frac{\sqrt{3}}{2}\mathrm{a}+\frac{\sqrt{3}}{2}\mathrm{b}\right)\mathrm{i}=2+\sqrt{3}\mathrm{i}$

Comparando as partes reais e as imaginárias:

$\left\{\begin{array}{l}-\frac{\mathrm{a}}{2}+\frac{\mathrm{b}}{2}=2\\ \frac{\sqrt{3}}{2}\mathrm{a}+\frac{\sqrt{3}}{2}\mathrm{b}=\sqrt{3 }\end{array}\right. \to \left\{\begin{array}{l}\mathrm{b}-\mathrm{a}=4\\ \mathrm{b}+\mathrm{a}=2\end{array}\right.$

Resolvendo o sistema, tem-se que: 

$\boxed{\mathrm{a}=-1 \mathrm{e} \mathrm{b}=3}$