De Moivre’s Theorem

De Moivre’s Theorem relates powers of complex numbers in polar form:

\( ( \cos \theta + i \sin \theta )^n = \cos(n\theta) + i \sin(n\theta) \), for all \( n \in \mathbb{Z} \)

If a complex number is written in polar form as:

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

Then its powers are given by:

\( z^n = r^n ( \cos(n\theta) + i \sin(n\theta) ) \)

Extension to Rational Exponents

De Moivre’s theorem also extends to fractional exponents, useful in extracting roots of complex numbers. For \( n \in \mathbb{Q} \), the principal root is:

\( z^{1/n} = r^{1/n} \left[ \cos\left(\frac{\theta + 2k\pi}{n}\right) + i \sin\left(\frac{\theta + 2k\pi}{n}\right) \right] \) where \( k = 0, 1, …, n-1 \)

This gives the n distinct roots of the complex number \( z \).

Proof by Induction (for \( n \in \mathbb{Z}^+ \))

Step 1: Base Case (n = 1)

\( (\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta \).  Holds true.

Step 2: Inductive Step

Assume the result is true for \( n = k \):

\( (\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta) \)

Now prove for \( n = k + 1 \):

\( (\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta) \)

Using the inductive hypothesis and product of trig identities:

$ \begin{align*} &= [\cos(k\theta) + i\sin(k\theta)](\cos\theta + i\sin\theta) \\ &= \cos(k\theta)\cos\theta – \sin(k\theta)\sin\theta + i[\cos(k\theta)\sin\theta + \sin(k\theta)\cos\theta] \\ &= \cos((k+1)\theta) + i\sin((k+1)\theta) \end{align*} $

 Therefore, De Moivre’s Theorem holds for all \( n \in \mathbb{Z}^+ \).

Note:

• De Moivre’s theorem simplifies computing powers and roots of complex numbers in polar form.
• The theorem is valid for all real numbers \( n \in \mathbb{R} \), though a full proof requires tools from calculus and analysis.
• When extended to fractional exponents, the result yields multiple roots spaced equally around the unit circle.