Esercizio. 
Calcolare
Calcolare
![\[\sqrt[3]{-i}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d4a2fd0a15cca4dab1512ad6811e9056_l3.png "Rendered by QuickLaTeX.com")
Soluzione
Avendo un numero complesso 
, le radici 
-esime di 
sono
![\[\sqrt[n]{z} = \sqrt[n]{\vert z \vert} \left(\cos \left(\frac{\alpha}{n} + \frac{2k\pi}{n} \right) + i \sin \left(\frac{\alpha}{n} + \frac{2k\pi}{n} \right)\right) \qquad k=0,1,\dots,n-1.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-2618caf6709126bc6404289f6cca20db_l3.png "Rendered by QuickLaTeX.com")
Nel nostro caso
![\[a = 01, \quad b=-1, \quad n=3 \qquad \mbox{e} \qquad k=0,1,2\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-7a7b50eab9b1d9663f4d5620c24bd07a_l3.png "Rendered by QuickLaTeX.com")
con
![\[\vert z \vert = \sqrt{0^2+(-1)^2} = 1 \qquad \mbox{e} \qquad \alpha= \dfrac{3\pi}{2}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-ceed8a844e6cba65317d1f0ddff86e2c_l3.png "Rendered by QuickLaTeX.com")
Da notare che il numero sotto radice ha parte reale pari a zero e parte immaginaria negativa, quindi era immediatamente possibile dedurre l’angolo 
pari a 
Applicando la formula abbiamo
![\[\begin{aligned} \sqrt{z} & = \left(\cos \left(\frac{3\pi}{2}\;\dfrac{1}{3} + \dfrac{2k\pi}{3} \right) + i \sin \left(\frac{3\pi}{2}\;\dfrac{1}{3} + \dfrac{2k\pi}{3} \right)\right) \qquad \qquad k=0,1,2\\\\ & = \left(\cos \left(\frac{\pi}{2}+ \dfrac{2k\pi}{3} \right) + i \sin \left(\frac{\pi}{2} + \dfrac{2k\pi}{3} \right)\right) \qquad \qquad k=0,1,2 \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-6ab90a5e088464d036d90c820461b371_l3.png "Rendered by QuickLaTeX.com")
per cui
![\[\begin{aligned} & k=0 \qquad \qquad \sqrt[3]{-i} = \cos \left(\frac{\pi}{2} \right) + i \sin \left(\frac{\pi}{2} \right) = i \\\\ & k=1 \qquad \qquad \sqrt[3]{-i} = \cos \left(\frac{\pi}{2}+ \dfrac{2\pi}{3} \right) + i \sin \left(\frac{\pi}{2} + \dfrac{2\pi}{3} \right) = \\\\ & \qquad \qquad \qquad \qquad = \cos \left(\frac{7\pi}{6} \right) + i \sin \left(\frac{7\pi}{6} \right) = \\\\ & \qquad \qquad \qquad \qquad = \cos \left( \pi + \frac{\pi}{6} \right) + i \sin \left(\pi + \frac{\pi}{6} \right) = - \dfrac{1}{2} (\sqrt{3}+i)\\\\ & k=2 \qquad \qquad \sqrt[3]{-i} = \cos \left(\frac{\pi}{2}+ \dfrac{4\pi}{3} \right) + i \sin \left(\frac{\pi}{2} + \dfrac{4\pi}{3} \right) = \\\\ & \qquad \qquad \qquad \qquad = \cos \left(\frac{11\pi}{6} \right) + i \sin \left(\frac{11\pi}{6}\right) =\\\\ & \qquad \qquad \qquad \qquad = \cos \left(-\frac{\pi}{6} \right) + i \sin \left(-\frac{\pi}{6}\right) = \dfrac{1}{2} (\sqrt{3}-i) \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-fd9798936c61a47094c6d74f57ca5e70_l3.png "Rendered by QuickLaTeX.com")
quindi
![\[\boxed{ \sqrt[3]{-i} = \dfrac{1}{2} (\sqrt{3}-i), \sqrt[3]{-i}=- \dfrac{1}{2} (\sqrt{3}+i), \sqrt[3]{-i} = i }\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-8b5cb8b7ab6c39f674acf34080ce1884_l3.png "Rendered by QuickLaTeX.com")
Fonte: Matematica.verde 3 – Zanichelli