De Moivre's Theorem for Integral and Rational Indices

We already know that multiplying two complex numbers written in the $cis$ form is easy on the angles: $cis\,\theta_1 \cdot cis\,\theta_2 = cis(\theta_1+\theta_2)$, because $\cos$ and $\sin$ satisfy the angle-addition formulas. Squaring a single $cis\,\theta$ is just a special case of this: $(cis\,\theta)^2 = cis\,2\theta$. The natural question is what happens for a general power — does $(\cos\theta+i\sin\theta)^n$ always collapse to $\cos n\theta + i\sin n\theta$, for any integer $n$, not just $n=2$? The answer is yes, and this fact is De Moivre's theorem, named after the French mathematician Abraham de Moivre.

Theorem (integral index). For any real $\theta$ and any integer $n$,

The cleanest way to see why this must be true is to build it up in three stages, matching the three flavours of integer: positive, zero, and negative.

Positive integers — induction. For $n=1$ the two sides are identical, so the base case is free. Assume the formula already holds for some positive integer $k$, i.e. $(\cos\theta+i\sin\theta)^k=\cos k\theta + i\sin k\theta$. Multiply both sides by one more factor of $(\cos\theta+i\sin\theta)$ and expand the right-hand product using $i^2=-1$; grouping real and imaginary parts turns the product into $\cos(k\theta+\theta) + i\sin(k\theta+\theta)$ purely by the ordinary cosine/sine addition formulas. That is exactly the statement for $n=k+1$. So the claim, once true for one positive integer, is automatically true for the next, and by induction it holds for every positive integer.

Zero. $(\cos\theta+i\sin\theta)^0 = 1 = \cos 0 + i \sin 0$, so the formula holds trivially at $n=0$.

Negative integers. Write $n=-m$ with $m$ a positive integer. Then $(\cos\theta+i\sin\theta)^{-m}$ is $1/(\cos\theta+i\sin\theta)^m$, and the positive-integer case already tells us the denominator is $\cos m\theta + i\sin m\theta$. Rationalising this fraction (multiply top and bottom by $\cos m\theta - i\sin m\theta$, and use $\cos^2+\sin^2=1$) leaves exactly $\cos(-m)\theta + i\sin(-m)\theta$, which is $\cos n\theta + i\sin n\theta$. So the negative case follows from the positive case for free.

Putting the three cases together proves the theorem for every integer $n$, positive, zero, or negative — nothing is assumed about the sign of $n$ anywhere the theorem is actually used.

It is standard shorthand to write $cis\,\theta$ for $\cos\theta+i\sin\theta$; in that notation De Moivre's theorem is simply $(cis\,\theta)^n = cis(n\theta)$. A few immediate consequences are worth keeping handy: $(\cos\theta - i\sin\theta)^n = \cos n\theta - i\sin n\theta$ (apply the theorem to $-\theta$, since $\cos(-\theta)=\cos\theta$ and $\sin(-\theta)=-\sin\theta$); and because $(\cos\theta+i\sin\theta)(\cos\theta-i\sin\theta)=1$, each of $\cos\theta+i\sin\theta$ and $\cos\theta-i\sin\theta$ is literally the reciprocal of the other.

Rational indices. The theorem as stated is only about integer powers, because 'raising to a rational power' isn't single-valued the way integer powers are — a rational power like $z^{1/2}$ genuinely has two possible values, and $z^{1/3}$ has three. What survives for a rational exponent $n=p/q$ (with $q>0$) is a weaker, one-valued statement: $\cos n\theta + i\sin n\theta$ is one of the values of $(\cos\theta+i\sin\theta)^n$. This follows by raising $\cos n\theta+i\sin n\theta$ to the power $q$: using the integral-index theorem twice, $(\cos n\theta+i\sin n\theta)^q = \cos(nq)\theta + i\sin(nq)\theta = \cos p\theta+i\sin p\theta = (\cos\theta+i\sin\theta)^p$, which exactly says $\cos n\theta+i\sin n\theta$ is a $q^{th}$ root of $(\cos\theta+i\sin\theta)^p$ — i.e. a legitimate value of $(\cos\theta+i\sin\theta)^{p/q}$. This rational-index version is what powers the root-finding in the next section.

Worked example. Simplify $\dfrac{(\cos\alpha+i\sin\alpha)^3}{(\sin\alpha - i\cos\alpha)^5}$. …