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como se hace la raiz cubica de 1 - i
[tex]\\z=1-i\\ \sqrt[n]{z}=\sqrt[n]{|z|}({\cos\frac{\varphi+2k\pi}{n}+i\sin \frac{\varphi+2k\pi}{n})}\\ |z|=\sqrt{1^2+(-1)^2}=\sqrt2\\ \varphi=arctg \frac{1}{-1}=arctg-1=-\frac{\pi}{4}\\ (\sqrt[3]{1-i})_0=\sqrt[6]{2}({\cos\frac{-\frac{\pi}{4}}{3}+i\sin \frac{-\frac{\pi}{4}}{3})}=\sqrt[6]{2}({\cos-\frac{\pi}{12}+i\sin -\frac{\pi}{12}})\\ [/tex]
[tex](\sqrt[3]{1-i})_1=\sqrt[6]{2}({\cos\frac{-\frac{\pi}{4}+2\pi}{3}+i\sin \frac{-\frac{\pi}{4}+2\pi}{3})}=\sqrt[6]{2}({\cos\frac{\frac{7\pi}{4}}{3}+i\sin \frac{\frac{7\pi}{4}}{3})}=\\\sqrt[6]{2}({\cos\frac{7\pi}{12}+i\sin \frac{7\pi}{12}})\\ (\sqrt[3]{1-i})_2=\sqrt[6]{2}({\cos\frac{-\frac{\pi}{4}+4\pi}{3}+i\sin \frac{-\frac{\pi}{4}+4\pi}{3})}=[/tex]
[tex]\sqrt[6]{2}({\cos\frac{\frac{15\pi}{4}}{3}+i\sin \frac{\frac{15\pi}{4}}{3})}=\\\sqrt[6]{2}({\cos\frac{5\pi}{4}+i\sin \frac{5\pi}{4}})\\ (\sqrt[3]{1-i})_3=\sqrt[6]{2}({\cos\frac{-\frac{\pi}{4}+6\pi}{3}+i\sin \frac{-\frac{\pi}{4}+6\pi}{3})}=\sqrt[6]{2}({\cos\frac{\frac{23\pi}{4}}{3}+i\sin \frac{\frac{23\pi}{4}}{3})}=\\\sqrt[6]{2}({\cos\frac{23\pi}{12}+i\sin \frac{23\pi}{12}})\\ [/tex]