Answer to Question #142635 in Complex Analysis for Desmond

Question #142635
Using De Moivre's theorem or otherwise
find the six roots of the equation z^6 +1 = 0
giving your answers in the form e^iw where w= angle teta
1
Expert's answer
2020-11-23T19:22:12-0500

Let us find the six roots of the equation "z^6=-1" using de Moivre's Theorem for fractional power:


"[r(\\cos\\theta+i\\sin\\theta)]^{\\frac{1}{n}}=\\sqrt[n]{r}(\\cos\\frac{\\theta+2\\pi k}{n}+i\\sin\\frac{\\theta+2\\pi k}{n})" for "k\\in\\{0,1,...,n-1\\}."


In trigonometric form "-1=\\cos\\pi+i\\sin\\pi", that is "r=1, \\theta=\\pi." Consequently, the formula


"z_k=\\cos\\frac{\\pi+2\\pi k}{6}+i\\sin\\frac{\\pi+2\\pi k}{6}=e^{\\frac{\\pi+2\\pi k}{6}i}" for "k\\in\\{0,1,2,3,4,5\\}" gives all six different roots.


Therefore, we have:


"z_0=e^{\\frac{\\pi}{6}i}", "z_1=e^{\\frac{3\\pi}{6}i}=e^{\\frac{\\pi}{2}i}", "z_2=e^{\\frac{5\\pi}{6}i}", "z_3=e^{\\frac{7\\pi}{6}i}", "z_4=e^{\\frac{9\\pi}{6}i}=e^{\\frac{3\\pi}{2}i}", "z_5=e^{\\frac{11\\pi}{6}i}."




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