De Moivre's Theorem (Rational Exponent)

For rational exponent p/q (in lowest terms), there are exactly q distinct values, obtained by taking k = 0, 1, 2, …, q−1. Unlike the integer case, the result is not unique — rational powers of a complex number are multi-valued.

For integer exponents, De Moivre's theorem gives one answer. For rational exponents $p/q$, it gives exactly $q$ distinct answers. This multi-valuedness is not an anomaly — it is the correct behaviour.

Setup

We want to find all values of $(\cos\theta + i\sin\theta)^{p/q}$ where $p/q$ is a fraction in lowest terms ($q > 0$).

Let $w = (\cos\theta + i\sin\theta)^{p/q}$. Then $w^q = (\cos\theta + i\sin\theta)^p$.

By De Moivre's theorem for integers:

$$w^q = \cos p\theta + i\sin p\theta = e^{ip\theta}$$

So we need all complex numbers $w$ satisfying $w^q = e^{ip\theta}$.

Finding All Solutions

Write $w = Re^{i\phi}$. Then $w^q = R^q e^{iq\phi}$. Equating with $e^{ip\theta}$:

$$R^q = 1 \implies R = 1$$ $$q\phi = p\theta + 2k\pi, \quad k \in \mathbb{Z} \implies \phi = \frac{p\theta + 2k\pi}{q}$$

So:

$$w = \cos\!\left(\frac{p\theta + 2k\pi}{q}\right) + i\sin\!\left(\frac{p\theta + 2k\pi}{q}\right)$$

Exactly q Distinct Values

As $k$ varies over all integers, the angle $\phi = (p\theta + 2k\pi)/q$ increases by $2\pi/q$ for each step of $k$. After $q$ steps, the angle increases by $2\pi$, returning to the same point.

So the distinct values correspond to $k = 0, 1, 2, \ldots, q-1$:

$$w_k = \cos\!\left(\frac{p\theta + 2k\pi}{q}\right) + i\sin\!\left(\frac{p\theta + 2k\pi}{q}\right), \quad k = 0, 1, \ldots, q-1$$

There are exactly $q$ distinct values, equally spaced at angular intervals of $2\pi/q$ on the unit circle.

Why Integer Exponents Give One Value

For $n \in \mathbb{Z}$, we need $w^1 = e^{in\theta}$, which has exactly $q = 1$ solution. The multi-valuedness only arises when the denominator $q > 1$.

The Contrast with Real Numbers

For real $x > 0$, $x^{p/q}$ has a unique positive real value by convention. For complex numbers, no such canonical choice exists — all $q$ values are geometrically equivalent, sitting at equal angles on the unit circle. Choosing one arbitrarily breaks the symmetry of the problem.