Estude! Com Prof. Caju

O resultado da potenciação do número complexo [tex3]\(\frac{1}{2}+i\frac{\sqrt{3}}{2}\)^{13}[/tex3] é:

a) 1
b) [tex3]\(\frac{\sqrt{3}}{2}+i\frac{1}{2}\)[/tex3]
c) [tex3]\(\frac{\sqrt{3}}{2}-i\frac{1}{2}\)[/tex3]
d) [tex3]\(\frac{1}{2}+i\frac{\sqrt{3}}{2}\)[/tex3]
e) [tex3]\(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\)[/tex3]

Admitindo-se o complexo:

[tex3]z=\frac{1}{2}+i\frac{\sqrt{3}}{2}[/tex3]

Temos que:

[tex3]a=\frac{1}{2}[/tex3]
[tex3]b=\frac{\sqrt{3}}{2}[/tex3]

Encontrando o afixo, temos:

[tex3]\varphi^2=a^2+b^2[/tex3]
[tex3]\varphi^2=\(\frac{1}{2}\)^2+\(\frac{\sqrt{3}}{2}\)^2[/tex3]
[tex3]\varphi^2=\frac{1}{4}+\frac{3}{4}[/tex3]
[tex3]\varphi^2=\frac{4}{4}[/tex3]
[tex3]\varphi^2=1[/tex3]
[tex3]\varphi=\sqrt{1}[/tex3]
[tex3]\varphi=1[/tex3]

Dos ângulos, temos:

[tex3]\cos \theta=\frac{a}{\varphi}=\frac{\frac{1}{2}}{1}=\frac{1}{2}[/tex3]
[tex3]\sen \theta=\frac{b}{\varphi}=\frac{\frac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3}}{2}[/tex3]

Logo:

[tex3]\theta=60[/tex3]

Na forma trigonométrica:

[tex3]z=1(\cos 60+i\sen60)[/tex3]

Elevando ao expoente desejado:

[tex3]z^{13}=1^{13}(\cos 13\cdot 60+i\sen13\cdot 60)[/tex3]
[tex3]z^{13}=\cos 780+i\sen780[/tex3]
[tex3]z^{13}=\cos 60+i\sen60[/tex3]n
[tex3]z^{13}=\frac{1}{2}+i\frac{\sqrt{3}}{2}[/tex3]