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Resposta:
Explicação passo a passo:
Seja Z = -2 + 2i
[tex]|Z|=\sqrt{a^2 + b^2} \\\\|Z|=\sqrt{(-2)^2+2^2}\\\\|Z| = \sqrt{4+4}\\\\|Z|=\sqrt{8}[/tex]
a < 0 e b > 0
[tex]cos\alpha =\frac{a}{|Z|} \\\\cos\alpha =-\frac{2}{2\sqrt{2} }\\\\cos\alpha =-\frac{1}{\sqrt{2}} =-\frac{\sqrt{2} }{2}\\\\sen\alpha =\frac{b}{|Z|} \\\\sen\alpha =\frac{2}{2\sqrt{2} }\\\\sen\alpha =\frac{1}{\sqrt{2} }=\frac{\sqrt{2} }{2}\\\\ \alpha =\pi -\frac{\pi }{4}=\frac{3\pi }{4} \\\\Z=|Z|(cos\alpha +isen\alpha )\\\\Z=2\sqrt{2} (cos\frac{3\pi }{4} +isen\frac{3\pi }{4})\\\\\sqrt[n]{Z}=\sqrt[n]{|Z|}~( cos \frac{\alpha +2k\pi }{n} +isen\frac{\alpha \\+2k\pi }{n}),~onde~k=0, 1 e2[/tex]
[tex]Sejam~~w_0,w_1~e~w_2~~as~raizes\\\\w_0=\sqrt[3]{2\sqrt{2} }(cos\frac{\frac{3\pi }{4} +2.0.\pi }{3}+isen\frac{\frac{3\pi }{4}+2.0.\pi }{3}) \\\\w_0=\sqrt[3]{2\sqrt{2} }(cos\frac{\pi }{4}+isen\frac{\pi }{4}) \\\\w_1=\sqrt[3]{2\sqrt{2} }(cos\frac{\frac{3\pi }{4}+2.1.\pi }{3} +isen\frac{\frac{3\pi }{4}+2.1.\pi }{3} )\\\\w_1=\sqrt[3]{2\sqrt{2} } (cos\frac{11\pi }{12 }+isen\frac{11\pi }{12} )\\\\w_2=\sqrt[3]{2\sqrt{2} }(cos\frac{\frac{3\pi }{4}+2.2.\pi }{3}+isen\frac{\frac{3\pi }{4}+2.2.\pi }{3} )[/tex]
[tex]w_2=\sqrt[3]{2\sqrt{2} }(\frac{19\pi }{12}+isen\frac{19\pi }{12} )[/tex]