De Moivre’s Theorem and Roots of Complex Numbers

De Moivre’s Theorem is a powerful tool in complex algebra that relates complex numbers in polar form to trigonometry. It provides an efficient method for finding powers and roots of complex numbers.

1. De Moivre’s Theorem for Integer Powers

De Moivre’s Theorem states that for any complex number \(z = r(\cos \theta + i \sin \theta) = r\text{cis}(\theta)\) and any integer \(n \in \mathbb{Z}\):

Key Properties:

• Magnitude: The magnitude of the result is the original magnitude raised to the power of \(n\) (\(|z^n| = |z|^n\)).
• Argument: The argument of the result is \(n\) times the original argument (\(\text{Arg}(z^n) = n\text{Arg}(z)\)).
• Negative Integers: The theorem holds for negative integers. Note that \(z^{-n} = \frac{1}{z^n}\).
• Principal Argument: After applying the theorem, the resulting argument \(n\theta\) may fall outside the principal range \((-\pi, \pi]\). It is standard practice to adjust the argument by adding or subtracting multiples of \(2\pi\) to return it to the principal range.

EXAM TIP: When raising a complex number in Cartesian form (\(a + bi\)) to a high power, always convert it to polar form (\(r\text{cis}\theta\)) first to apply De Moivre’s Theorem. This is much faster and less error-prone than binomial expansion.

2. Finding Roots of Complex Numbers

To solve the equation \(z^n = w\), where \(w\) is a known complex number, we use the inverse application of De Moivre’s Theorem.

The General Procedure

To find the \(n\)th roots of a complex number \(w = R\text{cis}(\phi)\):

1. Express \(w\) in general polar form:
   Represent \(w\) using a general argument to account for all possible roots:
   \$\(w = R\text{cis}(\phi + 2k\pi), \text{ where } k \in \mathbb{Z}\)\$
2. Set up the equation:
   Let \(z = r\text{cis}(\theta)\). Then \(z^n = r^n\text{cis}(n\theta)\).
   \$\(r^n\text{cis}(n\theta) = R\text{cis}(\phi + 2k\pi)\)\$
3. Solve for \(r\) and \(\theta\):
   • \(r^n = R \implies r = \sqrt[n]{R}\)
   • \(n\theta = \phi + 2k\pi \implies \theta = \frac{\phi + 2k\pi}{n}\)
4. Generate the \(n\) roots:
   Substitute \(k = 0, 1, 2, \dots, n-1\) to obtain \(n\) distinct solutions.

KEY TAKEAWAY: A complex equation of the form \(z^n = w\) will always have exactly \(n\) distinct roots in the complex plane.

3. Roots of Unity

The solutions to the equation \(z^n = 1\) are known as the \(n\)th roots of unity.

• Since \(1 = 1\text{cis}(0)\), the roots are given by:
  \$\(z_k = \text{cis}\left(\frac{2k\pi}{n}\right), \text{ for } k = 0, 1, \dots, n-1\)\$
• The roots of unity form a geometric sequence: \(1, \omega, \omega^2, \dots, \omega^{n-1}\), where \(\omega = \text{cis}\left(\frac{2\pi}{n}\right)\).

Geometric Properties of Roots

The \(n\)th roots of any complex number \(z^n = w\) exhibit specific geometric patterns:
1. Modulus: All roots have the same modulus \(r = \sqrt[n]{|w|}\), meaning they lie on a circle of radius \(r\) centered at the origin.
2. Spacing: The roots are equally spaced around the circle at intervals of \(\frac{2\pi}{n}\) radians.
3. Polygon: The roots form the vertices of a regular \(n\)-sided polygon inscribed in the circle.
4. Sum of Roots: The sum of the \(n\)th roots of any complex number is always zero: \(\sum_{k=0}^{n-1} z_k = 0\).

VCAA FOCUS: VCAA often asks students to plot these roots on an Argand diagram. Remember to clearly label the circle radius and ensure the angular spacing looks uniform.

4. Solving Polynomial Equations

De Moivre’s Theorem is frequently used in conjunction with the Fundamental Theorem of Algebra, which states that a polynomial of degree \(n\) has exactly \(n\) linear factors (and thus \(n\) roots) over the complex field \(\mathbb{C}\).

Solving \(z^n = (a+bi)\)

When the RHS is not in polar form:
1. Calculate \(R = \sqrt{a^2 + b^2}\).
2. Calculate \(\phi = \text{Arg}(a+bi)\).
3. Apply the roots formula: \(z = R^{1/n}\text{cis}\left(\frac{\phi + 2k\pi}{n}\right)\).

Example: Finding Cube Roots

To solve \(z^3 = 8i\):
1. \(8i = 8\text{cis}(\frac{\pi}{2})\).
2. \(z = \sqrt[3]{8}\text{cis}\left(\frac{\pi/2 + 2k\pi}{3}\right) = 2\text{cis}\left(\frac{\pi + 4k\pi}{6}\right)\).
3. \(k=0: z_0 = 2\text{cis}(\frac{\pi}{6})\)
4. \(k=1: z_1 = 2\text{cis}(\frac{5\pi}{6})\)
5. \(k=-1: z_2 = 2\text{cis}(-\frac{\pi}{2})\) (using \(k=-1\) often yields the principal argument directly).

COMMON MISTAKE: Forgetting to include the \(2k\pi\) term. If you only solve \(\frac{\phi}{n}\), you will only find one root (the principal root), missing the other \(n-1\) solutions required for a complete answer.

5. Summary of Applications

• Simplifying Expressions: Reducing terms like \((\cos \theta + i \sin \theta)^{10}\) to \(\cos 10\theta + i \sin 10\theta\).
• Trigonometric Identities: Expressing \(\cos(n\theta)\) and \(\sin(n\theta)\) in terms of powers of \(\cos \theta\) and \(\sin \theta\) by expanding \((\cos \theta + i \sin \theta)^n\) using the Binomial Theorem and equating real/imaginary parts.
• Factorisation: Factorising polynomials of the form \(z^n - a^n = 0\) into linear factors using the roots found via De Moivre’s Theorem.

STUDY HINT: Practice converting roots back into Cartesian form (\(x + iy\)). VCAA questions often specify the required form for the final answer, and being fluent in exact value triangles (for \(\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}\)) is essential.