Question

Determine as raízes cúbicas de -i. (Números Complexos - 2ª Fórmula de De Moivre).

Answer 1

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Raízes enésimas de um número complexo

$\large\boxed{\boxed{\boxed{\boxed{\sf w_k=\sqrt[\sf n]{\sf\rho}\bigg[cos\bigg(\dfrac{\theta+2k\pi}{n}\bigg)+i~sen\bigg(\dfrac{\theta+2k\pi}{n}\bigg)\bigg]}}}}$

$\sf z=-i\\\sf \rho=\sqrt{(-1)^2}=|-1|=1\\\sf sen(\theta)=\dfrac{-1}{1}\\\sf sen(\theta)=-1\implies \theta=arcsen(-1)=\dfrac{3\pi}{2}\\\sf w_k=\sqrt[\sf3]{\sf1}\bigg[cos\bigg(\dfrac{\frac{3\pi}{2}+2k\pi}{3}\bigg)+i~sen\bigg(\dfrac{\frac{3\pi}{2}+2k\pi}{3}\bigg)\bigg]\\\sf w_k=cos\bigg(\dfrac{\pi}{2}+\dfrac{2k\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi}{2}+\dfrac{2k\pi}{3}\bigg)$

$\sf se~ k=0:\\\sf w_0=cos\bigg(\dfrac{\pi}{2}\bigg)+i~sen\bigg(\dfrac{\pi}{2}\bigg)\\\sf w_0=0+i\cdot1\\\sf w_0=i$

$\sf se~k=1:\\\sf w_1=cos\bigg(\dfrac{\pi}{2}+\dfrac{2\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi}{2}+\dfrac{2\pi}{3}\bigg)\\\sf w_1=cos\bigg(\dfrac{3\pi+4\pi}{6}\bigg)+i~sen\bigg(\dfrac{3\pi+4\pi}{6}\bigg)\\\sf w_1=cos\bigg(\dfrac{7\pi}{6}\bigg)+i~sen\bigg(\dfrac{7\pi}{6}\bigg)\\\sf w_1=-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}~i$

$\sf se~k=2\\\sf w_2=cos\bigg(\dfrac{\pi}{2}+\dfrac{4\pi}{3}\bigg)+i~sen\bigg(\dfrac{\pi}{2}+\dfrac{4\pi}{3}\bigg)\\\sf w_2=cos\bigg(\dfrac{3\pi+8\pi}{6}\bigg)+i~sen\bigg(\dfrac{3\pi+8\pi}{6}\bigg)\\\sf w_2=cos\bigg(\dfrac{11\pi}{6}\bigg)+i~sen\bigg(\dfrac{11\pi}{6}\bigg)\\\sf w_2=\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}~i$

$\boxed{\begin{array}{l}\sf portanto~as~ra\acute izes~c\acute ubicas~do~n\acute umero~z=-i~s\tilde ao:\\\boxed{\boxed{\boxed{\boxed{\sf i}}}}~~~\boxed{\boxed{\boxed{\boxed{\sf-\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}~i}}}}~~\sf e~\boxed{\boxed{\boxed{\boxed{\sf\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}i}}}}\end{array}}\blue{\checkmark}$