Questão 5 - 1ª Fase - IME 2027

Note que:

$\mathrm{z}=\frac{1-2\mathrm{i}}{4-3\mathrm{i}}·\frac{4+3\mathrm{i}}{4+3\mathrm{i}}=\frac{2}{5}-\frac{\mathrm{i}}{5}$

Também:

$\frac{\overline{\mathrm{z}}-{\left|\mathrm{z}\right|}^2}{{\left|1-\mathrm{z}\right|}^2}=\frac{\left({\displaystyle \frac{2}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}\right)-\left({\left({\displaystyle \frac{2}{5}}\right)}^2+{\left(-{\displaystyle \frac{1}{5}}\right)}^2\right)}{{\left(-{\displaystyle \frac{3}{5}}\right)}^2+{\left({\displaystyle \frac{1}{5}}\right)}^2}=\frac{{\displaystyle \frac{2}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}-{\displaystyle \frac{1}{5}}}{{\displaystyle \frac{2}{5}}}=\frac{1}{2}+\frac{\mathrm{i}}{2}$

Além disso, da soma da PG:

${\mathrm{S}}_{\mathrm{n}}=\frac{\overline{\mathrm{z}}\left({\left(\overline{\mathrm{z}}\right)}^{\mathrm{n}}-1\right)}{\overline{\mathrm{z}}-1}=\frac{\left({\displaystyle \frac{2}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}\right)\left({\left({\displaystyle \frac{2}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}\right)}^{\mathrm{n}}-1\right)}{\left({\displaystyle \frac{2}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}\right)-1}=\frac{{\left({\displaystyle \frac{2}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}\right)}^{\mathrm{n}+1}}{-{\displaystyle \frac{3}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}}+\frac{1}{2}+\frac{\mathrm{i}}{2}$

Por fim, tem-se:

 

$$\begin{gathered} \left|{\mathrm{S}}_{\mathrm{n}}-\frac{\overline{\mathrm{z}}-{\left|\mathrm{z}\right|}^2}{{\left|1-\mathrm{z}\right|}^2}\right|<\frac{1}{2027} \\ \left|\frac{{\left({\displaystyle \frac{2}{5}+\frac{\mathrm{i}}{5}}\right)}^{\mathrm{n}+1}}{-{\displaystyle \frac{3}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}}+\frac{1}{2}+\frac{\mathrm{i}}{2}-\frac{1}{2}-\frac{\mathrm{i}}{2}\right|<\frac{1}{2027} \\ \left|\frac{{\left({\displaystyle \frac{2}{5}+\frac{\mathrm{i}}{5}}\right)}^{\mathrm{n}+1}}{-{\displaystyle \frac{3}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}}\right|<\frac{1}{2027} \\ \frac{\left|{\left({\displaystyle \frac{2}{5}+\frac{\mathrm{i}}{5}}\right)}^{\mathrm{n}+1}\right|}{\left|-{\displaystyle \frac{3}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}\right|}<\frac{1}{2027} \\ {\left|\frac{2}{5}+\frac{\mathrm{i}}{5}\right|}^{\mathrm{n}+1}<\frac{\left|-{\displaystyle \frac{3}{5}}+{\displaystyle \frac{\mathrm{i}}{5}}\right|}{2027} \\ {\left(\frac{1}{\sqrt{5}}\right)}^{\mathrm{n}+1}<\frac{\sqrt{2}}{2027\sqrt{5}} \\ \frac{1}{{\sqrt{5}}^{\mathrm{n}}}<\frac{\sqrt{2}}{2027}\to {\sqrt{5}}^{\mathrm{n}}>\frac{2027}{\sqrt{2}} \\ {5}^{\mathrm{n}}>\frac{{2027}^2}{2} \end{gathered}$$

Por inspeção, tem-se que o primeiro n inteiro que satisfaz a condição é $\boxed{\mathrm{n} = 10}$.