Introduction to De Moivre's Theorem
De Moivre's theorem is a powerful tool in complex analysis that connects trigonometry with complex numbers. It states that for any real number $\theta$ and any integer $n$:
$$(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)$$
This formula makes calculating powers and roots of complex numbers significantly easier by transforming algebraic exponentiation into trigonometric multiplication.
Finding the Cube Roots of Unity
The cube roots of unity are the solutions to the equation $z^3 = 1$.
Step 1: Represent 1 in Polar Form
We can write $1$ in complex polar form as: $$1 = \cos(0 + 2k\pi) + i \sin(0 + 2k\pi)$$ where $k$ is an integer.
Step 2: Apply De Moivre's Theorem
To find the cube roots, we raise both sides to the power of $1/3$: $$z = (\cos(2k\pi) + i \sin(2k\pi))^{1/3}$$ $$z = \cos\left(\frac{2k\pi}{3}\right) + i \sin\left(\frac{2k\pi}{3}\right)$$
Step 3: Solve for $k = 0, 1, 2$
- For $k=0$: $z_0 = \cos(0) + i \sin(0) = 1$
- For $k=1$: $z_1 = \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$
- For $k=2$: $z_2 = \cos\left(\frac{4\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$
Proving the Properties of $\omega$
Let $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. We are asked to show that $\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Calculation: $$\omega^2 = \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^2$$ Using the identity $(a+b)^2 = a^2 + 2ab + b^2$: $$\omega^2 = \left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right)\left(i\frac{\sqrt{3}}{2}\right) + \left(i\frac{\sqrt{3}}{2}\right)^2$$ $$\omega^2 = \frac{1}{4} - i\frac{\sqrt{3}}{2} - \frac{3}{4}$$ $$\omega^2 = -\frac{2}{4} - i\frac{\sqrt{3}}{2}$$ $$\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$$
This confirms the relationship between the two complex cube roots of unity.