91Ó°ÊÓ

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

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 $\sqrt{-}1$ .

Consider the equation $z=r×e^{\left({\mathrm{θ}}_i\right)}$

The magnitude of the complex number is r = 8

The argument of the complex number is$\mathrm{θ}=2\mathrm{π}$

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

Angle ${\mathrm{θ}}_k$is given as ${\mathrm{θ}}_k=\frac{2\mathrm{π}+2\mathrm{π}k}{n}$.

Find the different roots of the complex number.

Solve z and $\mathrm{θ}$for k = 0,1 .

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

Solve z and $\mathrm{θ}$ for k= 2.

$$\begin{gathered} {\mathrm{θ}}_2=2\mathrm{π} \\ {z}_2=2e^{\left(2\mathrm{π}\right)} \end{gathered}$$

Solve for $z_0$

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

For $z_1$

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

For $z_2$

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

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 $z_0+z_1+z_2=0$.