Question

Resolva a equação binômial
x³ + i = 0

Answer 1

$\displaystyle \sf \text{Radicia{\c c}{\~a}o de um complexo} : \\\\ \sqrt[\displaystyle \sf n]{\sf Z} = \sqrt[\displaystyle \sf n]{\sf |Z|} \cdot cis\left(\frac{arg(z)+2\cdot k\cdto \pi }{n}\right) \ , \ k = \{0,1,2, \ ... \ ,n-1 \}$

Temos :

$\displaystyle \sf x^3+i = 0 \\\\ x^3 = -i \\\\ x^3 = cis\left(\frac{-\pi }{2}+2\cdot k\cdot \pi \right) \ \ k \in \mathbb{Z} \\\\\\ \sqrt[3]{\sf x^3 } = cis\left( \frac{\displaystyle \frac{-\pi }{2}+2\cdot k\cdot \pi}{\displaystyle 3} \right) \\\\\\ x = cis\left(\frac{-\pi+4\cdot k\cdot\pi }{6}\right) \ \ ,\ \ k = \{0,1,2\}$

$\displaystyle \sf k = 0 \to x_ 1 = cis\left (\frac{-\pi }{6}\right ) = cos\left(\frac{-\pi}{6} \right) +i\cdot sen \left(\frac{-\pi}{6} \right) = \frac{\sqrt{3}}{2}-i\cdot \frac{1}{2} } \\\\\\ k = 1 \to x_ 2 = cis\left (\frac{-\pi +4\pi }{6}\right ) = cos\left(\frac{3\pi}{6} \right) +i\cdot sen \left(\frac{3\pi}{6} \right) = i$

$\displaystyle \sf k = 2 \to x_ 3 = cis\left (\frac{-\pi +4\cdot 2\pi }{6}\right ) = cos\left(\frac{7\pi}{6} \right) +i\cdot sen \left(\frac{7\pi}{6} \right) = \frac{-\sqrt{3}}{2}-i\cdot \frac{1}{2}$

Daí, temos as seguinte raízes :

$\boxed{ \ \begin{array}{I} \\ \displaystyle \sf x_ 1 = \frac{\sqrt{3}}{2}-i\cdot\frac{1}{2} \\\\\\ \sf x_2 = i\\\\\\ \displaystyle \sf x_ 3 = -\frac{\sqrt{3}}{2}-i\cdot\frac{1}{2} \\\\ \end{array}\ } \huge{\checkmark}$