Questão 6 - 1ª Fase - ITA 2027

Tem-se:

$\left|\mathrm{z}\right|={\left|\frac{3}{4}+\frac{\sqrt{3}}{4}\mathrm{i}\right|}^{\mathrm{m}} ={\left(\sqrt{{\left(\frac{3}{4}\right)}^2+{\left(\frac{\sqrt{3}}{4}\right)}^2}\right)}^{\mathrm{m}}={\left(\frac{\sqrt{3}}{2}\right)}^{\mathrm{m}}$

Assim:

${\left(\frac{\sqrt{3}}{2}\right)}^{\mathrm{m}}=\frac{81}{256}\Rightarrow {\left(\frac{\sqrt{3}}{2}\right)}^{\mathrm{m}}={\left(\frac{\sqrt{3}}{2}\right)}^8\Rightarrow \mathrm{m}=8$

Também:

$\left|\mathrm{w}\right|={\left|-\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}\mathrm{i}\right|}^{\mathrm{n}} ={\left(\sqrt{{\left(\frac{\sqrt{2}}{2}\right)}^2+{\left(\frac{\sqrt{6}}{2}\right)}^2}\right)}^{\mathrm{n}}={\left(\sqrt{2}\right)}^{\mathrm{n}}$

Assim:

${\left(\sqrt{2}\right)}^{\mathrm{n}}=16\Rightarrow \mathrm{n}=8$

Por fim:

$\arg \left(\mathrm{z}\right)=8·\arg \left(\frac{3}{4}+\frac{\sqrt{3}}{4}\mathrm{i}\right)=8·\frac{\mathrm{\pi }}{6}=\frac{4\mathrm{\pi }}{3}$

$\arg \left(\mathrm{w}\right)=8·\arg \left(-\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}\mathrm{i}\right)=8·\frac{2\mathrm{\pi }}{3}=\frac{16\mathrm{\pi }}{3}\equiv \frac{4\mathrm{\pi }}{3}$

Então:

$\boxed{\arg \left(\mathrm{zw}\right)=\arg \left(\mathrm{z}\right)+\arg \left(\mathrm{w}\right)=\frac{4\mathrm{\pi }}{3}+\frac{4\mathrm{\pi }}{3}=\frac{8\mathrm{\pi }}{3}\equiv \frac{2\mathrm{\pi }}{3}}$