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find all the complex cube roots of z=8. write each root in rectangular form.

 Aug 1, 2014

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 #1
avatar+4472 
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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.

 Aug 1, 2014
 #1
avatar+4472 
+5
Best Answer

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.

AzizHusain Aug 1, 2014

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