Question

Express the roots of the equation z3 − α3 = 0 in terms of α and w, where w is
a complex cube root of unity. Use your answer to find the roots of the following
equations in the form a + ib.

Answer 1

Given,

we have to find the roots of the equation, $z^3 -\alpha^3=0 \ in \ terms\ of \alpha \ and \ \omega .$

We know,

$z^3-\alpha^3 = (z-\alpha)(z^2 + \alpha^2 +z\alpha)=0 \ which \ implies\ z=\alpha\ (a\ root) \ and\ z^2+\alpha^2+z\alpha=0$

Now,

$z^2+\alpha^2+z\alpha=z^2 +2.z.\frac{\alpha}{2}+(\frac{\alpha}{2})^2+\alpha^2-(\frac{\alpha}{2})^2=0 \ implies\ (z+(\frac{\alpha}{2}))^2+(\frac{3\alpha}{4})^2=0$

$(z+(\frac{\alpha}{2}))=\frac{+}{-} \frac{i\sqrt{3}\alpha}{2}$

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:

$z=\alpha,\alpha$ $((\frac{-1}{2})+\frac{i\sqrt{3}}{2}) \ and\ \alpha((\frac{-1}{2})-\frac{i\sqrt{3}}{2})$.