Q30P

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

Verified

Answer

The sum of three cube roots is $z_{0} + z_{1} + z_{2} = 0$ .

1Step 1: Given Information

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

2Step 2: Definition of the complex number

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

$z = x + i y$

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

3Step 3: Finding the roots

Consider the equation $z = r \times e^{(θ_{i})}$

The magnitude of the complex number is r = 8

The argument of the complex number is $θ = 2 π$

Write the root in exponential form $z_{k} = r^{\frac{1}{n}} e^{(θ_{k} i)}$ .

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

Find the different roots of the complex number.

Solve z and $θ$ for k = 0,1 .

$θ_{0} = \frac{2 π}{3} z_{0} = 2 e^{(2 π / 3)} θ_{1} = \frac{4 π}{3} z_{1} = 2 e^{(4 π / 3)}$

Solve z and $θ$ for k = 2.

$θ_{2} = 2 π z_{2} = 2 e^{(2 π)}$

4Step 4: Solving the Cartesian form of root

Solve for $z_{0}$

$z_{0} = 2 [\cos (\frac{2 π}{3}) + i \sin (\frac{2 π}{3})] = 1 + i \sqrt{3}$

For $z_{1}$

$z_{1} = 2 [\cos (\frac{4 π}{3}) + i \sin (\frac{4 π}{3})] = - 1 - \sqrt{3 i}$

For $z_{2}$

$z_{2} = 2 [\cos (2 π) + i \sin (2 π)] = 2$

Add $z_{0}, z_{1}, z_{2}$ .

$z_{0} + z_{1} + z_{2} = - 1 - i \sqrt{3} - 1 + \sqrt{3 i} + 2 = 0$

Hence the solution is $z_{0} + z_{1} + z_{2} = 0$ .