Answer to Question #116900 in Algebra for kay

Question

8. Express the roots of the equation z^3 − α^3 = 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.
(a) z^3 − 27 = 0
(b) z^3 + 8 = 0
(c) z^3 − i = 0.

Answer 1

$z^3 −α^3 =0; \\ z^3-\alpha^3=(z-\alpha)(z^2+\alpha z+\alpha^2); \\ \implies (z-\alpha)(z^2+\alpha z+\alpha^2)=0; \\ \implies z-\alpha=0 \ or \ z^2+\alpha z+\alpha^2=0; \\ \implies z_1=\alpha; D=\alpha^2-4\alpha^2=3\alpha^2; \\ z_2=\dfrac{-\alpha+\sqrt{3\alpha^2}}{2}=\alpha \dfrac{-1+i\sqrt{3}}{2}; z_3=\dfrac{-\alpha-\sqrt{3\alpha^2}}{2}=\alpha \dfrac{-1-i\sqrt{3}}{2}$ .

Thus roots of given equation are: $\alpha, \alpha w , \alpha w^2$ .

a) Given equation is $z^3 - 27 =0$ ;

By comparing with above equation we have $\alpha^3 = 27 \implies \alpha = 3$

So, roots of given equation are $3, 3 (\dfrac{-1+i\sqrt{3}}{2}), 3(\dfrac{-1-i\sqrt{3}}{2})$ .

b) In this case $\alpha^3 = - 8 = (-2)^3 \implies \alpha = -2$

So, roots of given equation are $-2, -2 (\dfrac{-1+i\sqrt{3}}{2}), -2(\dfrac{-1-i\sqrt{3}}{2})$ .

c) In this case, $\alpha^3 = i = (-i)^3 \implies \alpha = -i$

So, roots of given equation are $-i, -i (\dfrac{-1+i\sqrt{3}}{2}), -i(\dfrac{-1-i\sqrt{3}}{2})$

Thus, roots in form of a+ib are $-i, \dfrac{\sqrt{3}+i}{2}, \dfrac{-\sqrt{3}+i}{2}$ .