Question

9. Seja
$z = - \sqrt{2 + i \sqrt[]{2} }$
determine:

z⁶
​

Answer 1

Resposta: $\color{green} \boxed{{ 4 - 10 \sqrt{2} \: i }}$

${ - \sqrt{2 + i \sqrt{2} } }^{ \: 6} \\ { - \sqrt[2 \div 2]{2 + i \sqrt{2} } }^{ \: 6 \div 2} \\ - \lgroup2 + i \sqrt{2} \rgroup {}^{3} \\ - \lgroup2 + \sqrt{2} \: i\rgroup {}^{3} \\ -\lgroup {2}^{3} + 3 \times {2}^{2} \sqrt{2} \: i + 3 \times 2 \times \lgroup \sqrt{2} \: i \rgroup {}^{2} + {\lgroup \sqrt{2} \: i \rgroup}^{3} \rgroup \\ - \lgroup8 + 3 \times 4 \sqrt{2} \: i + 3 \times 2 \times {2i}^{2} + 2 \sqrt{2} \: {i}^{3}\rgroup \\ - \lgroup8 + 12 \sqrt{2} \: i + 12 {i}^{2} + 2 \sqrt{2} \: {i}^{3} \rgroup \\ - \lgroup8 + 12 \sqrt{2} \: i + 12 \times ( - 1) + 2 \sqrt{2} \times ( - 1)\rgroup \\ - \lgroup8 + 12 \sqrt{2} \: i - 12 - 2 \sqrt{2} \: i\rgroup \\ - \lgroup - 4 + 12 \sqrt{2} \: i - 2 \sqrt{2} \: i\rgroup \\ - \lgroup - 4 + 10 \sqrt{2} \: i \rgroup \\ \color{green} \boxed{{ 4 - 10 \sqrt{2} \: i }}$

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