2010013407

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Project ID: 
2010013407
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Accepted: 
0
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1
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0
Dva kořeny rovnice \[ x^{3} + 1 - \mathrm{i} = 0 \] jsou \[ \begin{aligned}x_{1}& = \root{6}\of{2}\left (\cos \frac{\pi} {4} + \mathrm{i}\sin \frac{\pi} {4} \right ),& \\x_{2}& = \root{6}\of{2}\left (\cos \frac{11} {12}\pi + \mathrm{i}\sin \frac{11} {12}\pi \right ). \\ \end{aligned} \] Určete třetí kořen.
\(x_{3} = \root{6}\of{2}\left (\cos \frac{19} {12}\pi + \mathrm{i}\sin \frac{19} {12}\pi \right )\)
\(x_{3} = \root{6}\of{2}\left (\cos \frac{7} {12}\pi + \mathrm{i}\sin \frac{7} {12}\pi \right )\)
\(x_{3} = \root{6}\of{2}\left (\cos \frac{5} {12}\pi + \mathrm{i}\sin \frac{5} {12}\pi \right )\)
\(x_{3} = \root{6}\of{2}\left (\cos \frac{13} {12}\pi + \mathrm{i}\sin \frac{13} {12}\pi \right )\)