Question

Show that the sum of the three cube roots of 8is zero.

Short answer

The sum of three cube roots is$z_0+z_1+z_2=0$ .

Step-by-step solution

Given Information

To prove that the sum of the three cube roots of 8 is zero.

Definition of the complex number

Complex numbers possess real numbers and imaginary numbers; a complex can be written in the form of:

$z=x+iy$

Here x and y are real numbers, and i is the imaginary number which is known as iota, whose value is $\sqrt{-}1$ .

Finding the roots

Consider the equation $z=r\times e^{\left({\theta }_i\right)}$

The magnitude of the complex number is r = 8

The argument of the complex number is$\theta =2\pi$

Write the root in exponential form $z_k=r^{\frac{1}{n}}{e}^{\left({\theta }_{k}i\right)}$.

Angle ${\theta }_k$is given as ${\theta }_k=\frac{2\pi +2\pi k}{n}$.

Find the different roots of the complex number.

Solve z and $\theta$for k = 0,1 .

$$\begin{gathered} {\theta }_0=\frac{2\pi }{3} \\ {z}_0=2e^{\left(2\pi /3\right)} \\ {\theta }_1=\frac{4\pi }{3} \\ {z}_1=2e^{\left(4\pi /3\right)} \end{gathered}$$

Solve z and $\theta$ for k= 2.

$$\begin{gathered} {\theta }_2=2\pi \\ {z}_2=2e^{\left(2\pi \right)} \end{gathered}$$

Solving the Cartesian form of root

Solve for $z_0$

$$\begin{gathered} z_0=2\left[\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\right] \\ =1+\mathrm{i}\sqrt{3} \end{gathered}$$

For $z_1$

$$\begin{gathered} z_1=2\left[\cos \left(\frac{4\pi }{3}\right)+i\sin \left(\frac{4\pi }{3}\right)\right] \\ =-1-\sqrt{3i} \end{gathered}$$

For $z_2$

$$\begin{gathered} z_2=2\left[\cos \left(2\pi \right)+i\sin \left(2\pi \right)\right] \\ =2 \end{gathered}$$

Add $z_0,z_1,z_2$.

$$\begin{gathered} z_0+z_1+z_2=-1-i\sqrt{3}-1+\sqrt{3i}+2 \\ =0 \end{gathered}$$

Hence the solution is $z_0+z_1+z_2=0$.