Given,
we have to find the roots of the equation, "z^3 -\\alpha^3=0 \\ in \\ terms\\ of \\alpha \\ and \\ \\omega ."
We know,
"z^3-\\alpha^3 = (z-\\alpha)(z^2 + \\alpha^2 +z\\alpha)=0\n\\ which \\ implies\\ z=\\alpha\\ (a\\ root) \\ and\\ z^2+\\alpha^2+z\\alpha=0"
Now,
"z^2+\\alpha^2+z\\alpha=z^2 +2.z.\\frac{\\alpha}{2}+(\\frac{\\alpha}{2})^2+\\alpha^2-(\\frac{\\alpha}{2})^2=0\n\\ implies\\ (z+(\\frac{\\alpha}{2}))^2+(\\frac{3\\alpha}{4})^2=0"
"(z+(\\frac{\\alpha}{2}))=\\frac{+}{-} \\frac{i\\sqrt{3}\\alpha}{2}"
or, "z=-(\\frac{\\alpha}{2})\\frac{+}{-}\\frac{i\\sqrt{3}\\alpha}{2}"
or, "z=\\alpha((\\frac{-1}{2})\\frac{+}{-}\\frac{i\\sqrt{3}}{2})"
or, "z=\\alpha\\omega\\ and\\ z=\\alpha\\omega^2\\ where \\ \\omega=((\\frac{-1}{2})+\\frac{i\\sqrt{3}}{2}) \\ and \\ \\omega^2=((\\frac{-1}{2})-\\frac{i\\sqrt{3}}{2})"
Thus, the three roots of the above equation are:
in terms of "\\omega" :
"z=\\alpha,\\alpha\\omega\\ and \\ \\alpha\\omega^2"
in terms of a+ib:
"z=\\alpha,\\alpha" "((\\frac{-1}{2})+\\frac{i\\sqrt{3}}{2}) \\ and\\ \\alpha((\\frac{-1}{2})-\\frac{i\\sqrt{3}}{2})".
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