Question

Considere o número complexo z = cos π/6 + i sen π/6. O valor de Z^3
+ Z^6
+
Z
^12 é:​

Answer 1

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Potenciação de números complexos

$\huge\boxed{\boxed{\boxed{\boxed{\sf z^n=\rho^n[cos(n\theta)+i~sen(n\theta)]}}}}$

$\sf z=cos\bigg(\dfrac{\pi}{6}\bigg)+i~sen\bigg(\dfrac{\pi}{6}\bigg)\\\sf\rho=1\\\sf\theta=\dfrac{\pi}{6}$

$\\\underline{\rm c\acute alculos~auxiliares:}\\\sf z^3=1^3\cdot\bigg[cos\bigg(\backslash\!\!\!3\cdot\dfrac{\pi}{\backslash\!\!\!6_2}\bigg)+i~sen\bigg(\backslash\!\!\!3\cdot\dfrac{\pi}{\backslash\!\!\!6_2}\bigg)\bigg]\\\sf z^3=1\bigg[cos\bigg(\dfrac{\pi}{2}\bigg)+i~sen\bigg(\dfrac{\pi}{2}\bigg)\bigg]=1\cdot i=i\\\sf z^6=(z^3)^2=(i)^2=-1\\\sf z^{12}=(z^6)^2=(-1)^2=1\\\sf z^3+z^6+z^{12}=i+(-1)+1\\\sf z^3+z^6+z^{12}=i-\backslash\!\!\!1+\backslash\!\!\!1$

$\Huge\boxed{\boxed{\boxed{\boxed{\sf z^3+z^6+z^{12}=i}}}}\blue{\checkmark}$