Answer to Question #117237 in Algebra for N kay

Question

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

Solution.

$z3 −α3 =0;$

$z^3-\alpha^3=(z-\alpha)(z^2+\alpha z+\alpha^2);$

$(z-\alpha)(z^2+\alpha z+\alpha^2)=0;$

$z−α=0 orz^2+\alpha z+\alpha^2=0;$

$z_1=\alpha;$ $D=\alpha^2-4\alpha^2=3\alpha^2$ ;

$z_2=\dfrac{-\alpha+\sqrt{3\alpha^2}}{2}=\alpha \dfrac{-1+i\sqrt{3}}{2};$

$z_3=\dfrac{-\alpha-\sqrt{3\alpha^2}}{2}=\alpha \dfrac{-1-i\sqrt{3}}{2};$

Cube Root of Unity Value:

$ω1 ​=1 - real;$

$\omega_2= \dfrac{-1+i\sqrt{3}}{2}$ - complex;

$\omega_3= \dfrac{-1-i\sqrt{3}}{2}$ - complex;

$​z_1=\alpha\sdot\omega_1=\alpha\sdot1=\alpha;$

$z_2=\alpha\sdot\omega_2=\alpha \dfrac{-1+i\sqrt{3}}{2}=-\dfrac{1}{2}\alpha+i\dfrac{\alpha\sqrt{3}}{2};$

$z_3=-\dfrac{1}{2}\alpha-i\dfrac{\alpha\sqrt{3}}{2};$

a) $z^3-27=0;$

$z_1=3;$

$z_2=-\dfrac{1}{2}\sdot3+i\dfrac{3\sqrt{3}}{2}=-\dfrac{3}{2}+i\dfrac{3\sqrt{3}}{2};$

$z_3=-\dfrac{1}{2}\sdot3-i\dfrac{3\sqrt{3}}{2}=-\dfrac{3}{2}-i\dfrac{3\sqrt{3}}{2};$

b)$z^3+8=0;$

$z_1=-2;$

$z_2=-\dfrac{1}{2}\sdot(-2)+i\dfrac{-2\sqrt{3}}{2}=-\dfrac{3}{2}-i\dfrac{2\sqrt{3}}{2};$

$z_3=-\dfrac{1}{2}\sdot(-2)-i\dfrac{-2\sqrt{3}}{2}=-\dfrac{3}{2}+i\dfrac{2\sqrt{3}}{2};$

c)$z^3-i=0;$

$z_1=i;$

$z_2=-\dfrac{1}{2}i+i\dfrac{i\sqrt{3}}{2}=-\dfrac{1}{2}i-\dfrac{\sqrt{3}}{2};$

$z_3=-\dfrac{1}{2}i-i\dfrac{i\sqrt{3}}{2}=-\dfrac{1}{2}i+\dfrac{\sqrt{3}}{2};$

Answer: a) $z_1=3;$

$z_2=-\dfrac{3}{2}+i\dfrac{3\sqrt{3}}{2};$

$z_3=-\dfrac{3}{2}-i\dfrac{3\sqrt{3}}{2};$

b)$z_1=-2;$

$z_2=-\dfrac{3}{2}-i\dfrac{2\sqrt{3}}{2};$

$z_3=-\dfrac{3}{2}+i\dfrac{2\sqrt{3}}{2};$

c)$z_1=i;$

$z_2=-\dfrac{1}{2}i-\dfrac{\sqrt{3}}{2};$

$z_3=-\dfrac{1}{2}i+\dfrac{\sqrt{3}}{2}.$