find all the complex cube roots of z=8. write each root in rectangular form.
+4473 cube root of z = 8
$$z^{3} = 8$$
$$z^{3} - 8 =0$$
$$(z-2)(z^{2}+2z+4)= 0$$
One solution is z - 2 = 0 --> z = 2.
For the other polynomial, let's use the quadratic formula where a = 1, b = 2, & c = 4: $$\begin{array}{*{20}c} {z = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{}}} \\ \end{array}$$
$$\begin{array}{*{20}c} {z = \frac{{ - 2 \pm \sqrt {4 - 16} }}{{2}}} & {{\rm{}}} \\ \end{array}$$
$$\begin{array}{*{20}c} {z = \frac{{-2 \pm \sqrt{-12} }}{{2}}} & {{\rm{}}} \\ \end{array}$$
We get z = -1 + 1.732050808i & -1 - 1.732050808i
The three complex roots are z = 2, z = -1 + 1.732050808i, & z = -1 - 1.732050808i.
+4473 cube root of z = 8
$$z^{3} = 8$$
$$z^{3} - 8 =0$$
$$(z-2)(z^{2}+2z+4)= 0$$
One solution is z - 2 = 0 --> z = 2.
For the other polynomial, let's use the quadratic formula where a = 1, b = 2, & c = 4: $$\begin{array}{*{20}c} {z = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{}}} \\ \end{array}$$
$$\begin{array}{*{20}c} {z = \frac{{ - 2 \pm \sqrt {4 - 16} }}{{2}}} & {{\rm{}}} \\ \end{array}$$
$$\begin{array}{*{20}c} {z = \frac{{-2 \pm \sqrt{-12} }}{{2}}} & {{\rm{}}} \\ \end{array}$$
We get z = -1 + 1.732050808i & -1 - 1.732050808i
The three complex roots are z = 2, z = -1 + 1.732050808i, & z = -1 - 1.732050808i.
