Understanding De Moivre's Theorem
De Moivre's Theorem is a powerful tool in complex analysis that links complex numbers and trigonometry. It states that for any real number $x$ and integer $n$:
$$(\cos x + i \sin x)^n = \cos(nx) + i \sin(nx)$$
This theorem is particularly useful for finding powers and roots of complex numbers, as it allows us to perform exponentiation and root extraction in polar form.
Finding the Square Roots of $2 + 2\sqrt{3}i$
To find the square roots of a complex number $z = 2 + 2\sqrt{3}i$, we first convert it to its polar form, $z = r(\cos \theta + i \sin \theta)$.
Step 1: Calculate Modulus and Argument
Modulus ($r$): $$r = |z| = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 4(3)} = \sqrt{4 + 12} = \sqrt{16} = 4$$
Argument ($\theta$): Since the number is in the first quadrant, $\theta = \arctan(\frac{2\sqrt{3}}{2}) = \arctan(\sqrt{3}) = 60^\circ$ or $\frac{\pi}{3}$ radians.
So, $z = 4(\cos 60^\circ + i \sin 60^\circ)$.
Step 2: Apply De Moivre's Theorem for Roots
The $n$-th roots of $z$ are given by: $$z^{1/n} = r^{1/n} \left[ \cos\left(\frac{\theta + 2k\pi}{n}\right) + i \sin\left(\frac{\theta + 2k\pi}{n}\right) \right]$$ where $k = 0, 1, 2, ..., n-1$. Here, $n=2$, so $k=0, 1$.
Step 3: Solve for $k=0$ and $k=1$
For $k = 0$: $$z_1 = 4^{1/2} \left[ \cos\left(\frac{60^\circ}{2}\right) + i \sin\left(\frac{60^\circ}{2}\right) \right] = 2(\cos 30^\circ + i \sin 30^\circ)$$ $$z_1 = 2\left(\frac{\sqrt{3}}{2} + i \frac{1}{2}\right) = \sqrt{3} + i$$
For $k = 1$: $$z_2 = 4^{1/2} \left[ \cos\left(\frac{60^\circ + 360^\circ}{2}\right) + i \sin\left(\frac{60^\circ + 360^\circ}{2}\right) \right] = 2(\cos 210^\circ + i \sin 210^\circ)$$ $$z_2 = 2\left(-\frac{\sqrt{3}}{2} - i \frac{1}{2}\right) = -\sqrt{3} - i$$
Conclusion
The square roots of $2 + 2\sqrt{3}i$ are $\pm(\sqrt{3} + i)$. De Moivre's theorem turns a complex algebraic problem into a straightforward geometric one by utilizing the rotation properties of complex numbers.