91Ó°ÊÓ

Cubing a number means raising it to the power of three. This is especially straightforward for real numbers. For example, the cube of 2 is calculated by multiplying 2 by itself three times:

• \( 2 imes 2 imes 2 = 2^3 = 8 \)

When dealing with complex numbers, the process requires a bit more thought. Complex numbers have both a real and an imaginary part, and cubing them can involve more than straightforward multiplication. The goal is to simplify the expression to a form that still represents a complex number, but now cubed.
There's often a necessity to convert complex numbers into another form, like polar form, to make operations such as cubing easier to handle.

The polar form is a way of representing complex numbers, which we often use for multiplication and exponentiation. Instead of using a standard format like \( a + bi \), polar form expresses complex numbers in terms of a magnitude (or modulus) and an angle (or argument):

• \( z = r (\cos \theta + i \sin \theta) \)

where \( r \) is the distance from the origin to the point \( (a, b) \) in the complex plane, given by \( r = \sqrt{a^2 + b^2} \), and \( \theta \) is the angle formed with the positive real axis, found using trigonometric relationships.
In the exercise, for instance, the number \( -1 + \sqrt{3}i \) converts to polar form with \( r = 2 \) and \( \theta = \frac{2\pi}{3} \). This transformation simplifies operations such as cubing, enabling us to apply formulas that are more manageable in polar coordinates.

De Moivre's Theorem provides a useful formula for raising complex numbers in polar form to a power. It states:

• If \( z = r (\cos \theta + i \sin \theta) \), then \( z^n = r^n (\cos(n\theta) + i \sin(n\theta)) \).

This theorem simplifies the process of cubing complex numbers. When given a complex number in polar form, such as \( -1 + \sqrt{3}i \), you can determine its cube by raising the modulus \( r \) to the power of 3 and multiplying the angle \( \theta \) by 3.
In our exercise, we used De Moivre's Theorem to turn a previously challenging problem into a more manageable one. Cubing the unit circle angle (\( 2\pi \)) brings us back to a whole number due to periodicity, simplifying operations significantly and leading to our answer.