Proszę o sprawdzenie:
\(\displaystyle{ \sqrt[5]{-1- \sqrt{3}i }}\)
\(\displaystyle{ z=-1- \sqrt{3}i}\)
\(\displaystyle{ \left|z \right|= \sqrt{4}=2}\)
\(\displaystyle{ cos \phi=- \frac{1}{2}}\)
\(\displaystyle{ sin \phi=- \frac{ \sqrt{3} }{2} \Rightarrow \phi= \frac{4\pi}{3}}\)
\(\displaystyle{ z_{0}= \sqrt[5]{2}(cos \frac{4\pi}{15} + isin \frac{4\pi}{15})}\)
\(\displaystyle{ z_{1}= \sqrt[5]{2}(cos \frac{10\pi}{15} + isin \frac{10\pi}{15})}\)
\(\displaystyle{ z_{2}= \sqrt[5]{2}(cos \frac{16\pi}{15} + isin \frac{16\pi}{15})}\)
\(\displaystyle{ z_{3}= \sqrt[5]{2}(cos \frac{22\pi}{15} + isin \frac{22\pi}{15})}\)
\(\displaystyle{ z_{4}= \sqrt[5]{2}(cos \frac{28\pi}{15} + isin \frac{28\pi}{15})}\)