Question

The three cube roots of \(+1\) are often called \(1, \omega\), and \(\omega^{2}\), Show that this is reasonable, that is, show that the cube roots of \(+1\) are \(+1\) and two other numbers, each of which is the square of the other.

Short answer

The cube roots of 1 are 1, \(\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\), and \(\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\).

Step-by-step solution

- Understanding cube roots

The cube roots of a number are values that, when raised to the power of 3, equal that number. For instance, if we want to find all the cube roots of 1, we need to solve the equation: \[x^3 = 1\]

- Consider known root

The first obvious root of the equation \(x^3 = 1\) is \(x = 1\). This is because \(1^3 = 1\).

- Complex roots using polar form

Any complex number can be written in the polar form as \(re^{iθ}\). Recall that 1 can be written as \[ 1 = e^{i \times 0}, e^{i \times 2\pi} \]

- Finding additional cube roots

To find other roots, express 1 using different multiples of \(2\pi\). We need values of \(\theta\) such that when divided by 3, results in the argument within one full rotation cycle \[ e^{i \times\frac{2\pi k}{3}} \text{ for } k = 0, 1, 2 \]

- Solve for \(k=1\)

When \(k = 1\): \[ e^{i \times \frac{2\pi \times 1}{3}} = e^{i \times \frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\] This is our second root \(\omega\).

- Solve for \(k=2\)

When \(k = 2\): \[ e^{i \times \frac{2\pi \times 2}{3}} = e^{i \times \frac{4\pi}{3}} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\] This is our third root \(\omega^2\).

- Verify relationship

To verify the relationship between \(\omega\) and \(\omega^2\), notice that: \[ \omega^3 = ( \omega) \times ( \omega) \times ( \omega) = 1 \] and \[ \omega^2 = (\omega)^2 \]

- Conclusion

Hence, the three cube roots of 1 are: 1, \(\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\), and \(\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\), where \(\omega\) and \(\omega^2\) are the complex conjugates of each other, and \(\omega^2\) is the square of \(\omega\).

Key concepts

complex numbers

Complex numbers are a type of number in the form of \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, with the property that \(i^2 = −1\). They are very useful in engineering, physics, and mathematics since they can represent two-dimensional quantities.
For instance, consider the cube roots of one. There are three complex numbers that satisfy this property:

• 1
• \(\frac{-1}{2} + i \frac{\text{√3}}{2}\) also known as \(\text{ω} \)
• \(\frac{-1}{2} - i \frac{\text{√3}}{2}\) also known as \( \text{ω}^2 \)

These roots lie on the complex plane and form an equilateral triangle. This connection works nicely with coordinates representing complex numbers. The geometric representation helps in understanding their behaviors and relationships.

polar form

The Polar form of a complex number is a way of expressing a complex number using the modulus (magnitude) and the angle (argument) relative to the positive real axis. Any complex number \(z = a + bi\) can be written in polar form as \(z = re^{iθ}\), where \(r = \text{√(a² + b²)}\) is the modulus and \(θ = \text{tan}^{-1}(b/a)\) is the argument.
The primary benefit of polar form is its usefulness in multiplying and dividing complex numbers. For example, to find the cube roots of 1, we write 1 as

• \(e^{i \times 0} \)
• \( e^{i \times 2π} \)

Then, solve the equation \(x^3 = 1\) by considering the multiples of \(2π\). This leads to values of \(θ\) that, when divided by 3, produce the arguments of cube roots within one full rotation:

• \( e^{i \times\frac{2πk}{3}}\) for \( k = 0, 1, 2 \)

Here, we obtain all three necessary cube roots.

Euler's formula

Euler's formula states that for any real number \(θ\), \( e^{iθ} = \text{cos}θ + i\text{sin}θ \). This formula beautifully connects complex exponential functions with trigonometric functions.
To see why Euler’s formula helps us find cube roots, let’s use it to express our solutions from earlier steps.

• \( e^{i \times \frac{0}{3}} = e^0 = 1\)
• \( e^{i \times \frac{2\text{π}}{3}} = \text{cos}(\frac{2\text{π}}{3}) + i \text{sin}(\frac{2\text{π}}{3}) = -\frac{1}{2} + i \frac{\text{√3}}{2}\)
• \( e^{i \times \frac{4\text{π}}{3}} = \text{cos}(\frac{4\text{π}}{3}) + i \text{sin}(\frac{4\text{π}}{3}) = -\frac{1}{2} - i \frac{\text{√3}}{2}\)

Euler’s formula simplifies root finding by transforming trigonometric operations into simpler exponential terms. It makes our computations elegant and easier to understand.