We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
1827
1
avatar

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

 Aug 1, 2014

Best Answer 

 #1
avatar+4472 
+5

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

23 Online Users

avatar
avatar
avatar
avatar