Questão 4 - 1ª Fase - IME 2026

Se $\left|\arg \left(\frac{{\mathrm{z}}_3-{\mathrm{z}}_2}{{\mathrm{z}}_1}\right)\right|=\frac{\mathrm{\pi }}{2}$, então $\frac{{\mathrm{z}}_3-{\mathrm{z}}_2}{{\mathrm{z}}_1}$ é imaginário puro, ou seja, $\frac{{\mathrm{z}}_3-{\mathrm{z}}_2}{{\mathrm{z}}_1}=\mathrm{ki}, \mathrm{k}\in {\mathrm{ℝ}}^{*}$.

Assim:

$\frac{{\mathrm{z}}_3-{\mathrm{z}}_2}{{\mathrm{z}}_1}=\mathrm{ki}\to \left|\frac{{\mathrm{z}}_3-{\mathrm{z}}_2}{{\mathrm{z}}_1}\right|=\left|\mathrm{ki}\right|\to \frac{\left|{\mathrm{z}}_3-{\mathrm{z}}_2\right|}{\left|{\mathrm{z}}_1\right|}=\left|\mathrm{k}\right|\left|\mathrm{i}\right|$

Desse modo:

$\frac{5}{5}=\left|\mathrm{k}\right|·1\to \mathrm{k}=\pm 1$

Analisando os dois casos separadamente, tem-se:

Caso 1: k = 1

$\frac{{\mathrm{z}}_3-{\mathrm{z}}_2}{{\mathrm{z}}_1}=1·\mathrm{i}\to {\mathrm{z}}_3={\mathrm{z}}_2+{\mathrm{iz}}_1\to {\mathrm{z}}_3=\left(12+5\mathrm{i}\right)+\mathrm{i}\left(3+4\mathrm{i}\right)$

${\mathrm{z}}_3=8+8\mathrm{i}$

$\boxed{{\mathrm{z}}_3·\overline{{\mathrm{z}}_3}={\left|{\mathrm{z}}_3\right|}^2={\left(\sqrt{8^2+8^2}\right)}^2=8^2+8^2=128}$

Caso 2: $\mathrm{k} = -1$

$\frac{{\mathrm{z}}_3-{\mathrm{z}}_2}{{\mathrm{z}}_1}=\left(-1\right)·\mathrm{i}\to {\mathrm{z}}_3={\mathrm{z}}_2-{\mathrm{iz}}_1\to {\mathrm{z}}_3=\left(12+5\mathrm{i}\right)-\mathrm{i}\left(3+4\mathrm{i}\right)$

${\mathrm{z}}_3=16+2\mathrm{i}$

$\boxed{{\mathrm{z}}_3·\overline{{\mathrm{z}}_3}={\left|{\mathrm{z}}_3\right|}^2={\left(\sqrt{{16}^2+2^2}\right)}^2={16}^2+2^2=260}$

 

Assim, o valor máximo é $\boxed{{\mathrm{z}}_3·\overline{{\mathrm{z}}_3}=260}$.