Proof of De Moivre's Theorem by induction

Proof of De Moivre's Theorem by induction

IB Year 6 | Grade 12

Let \(z=r(\cos\theta+i\sin\theta)\), with \(r>0\). We first establish by mathematical induction the statement \(P_n\) for non-negative integers:

\[z^n=r^n[\cos(n\theta)+i\sin(n\theta)],\qquad n\in\mathbb Z_{\ge0}.\]

The proof uses a base case at \(0\), followed by the implication from \(k\) to \(k+1\). The assumption \(r>0\) ensures \(z\ne0\) when zero or negative powers are used.

Base case: \(n = 0\).

\(z^0 = 1\) by definition.

\(r^0[\cos(0 \cdot \theta) + i\sin(0 \cdot \theta)] = 1\).

Thus, \(P_0\) is true.

Inductive step.

Assume that the statement \(P_k\) holds for some integer \(k \ge 0\), i.e.,

\[z^k = r^k[\cos(k\theta) + i\sin(k\theta)].\]

We want to prove that \(P_{k+1}\) is true, i.e.,

\[z^{k+1} = r^{k+1}[\cos((k+1)\theta) + i\sin((k+1)\theta)].\]

The proof is straightforward from our earlier result that

\[z_1z_2 = r_1r_2[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)].\]

Specifically, we have

\[\begin{aligned}z^{k+1}&=z^kz\\&=r^k[\cos(k\theta)+i\sin(k\theta)]r[\cos\theta+i\sin\theta]\\&=r^{k+1}[\cos((k+1)\theta)+i\sin((k+1)\theta)].\end{aligned}\]

and hence \(P_{k+1}\) is true.

Since \(P_0\) is true and \(P_k\Rightarrow P_{k+1}\) for every integer \(k\ge0\), mathematical induction proves \(P_n\) for every integer \(n\ge0\).

We now turn to the case when \(n\) is a negative integer. It follows immediately that \(m = -n\) is a positive integer.

So, by the result of reciprocation and by \(P_m\), we have

\[\begin{aligned}z^n&=\frac1{z^m}\\&=r^{-m}[\cos(m\theta)-i\sin(m\theta)]\\&=r^n[\cos(-n\theta)-i\sin(-n\theta)]\\&=r^n[\cos(n\theta)+i\sin(n\theta)].\end{aligned}\]

Thus, the result also holds for negative integers \(n\).

All in all, if \(z=r(\cos\theta+i\sin\theta)\) with \(r>0\), then

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

for all \(n \in \mathbb{Z}\).

In particular, when \(r = 1\), it follows that

\[(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta).\]

This particular result is known as de Moivre’s Theorem due to the French mathematician Abraham de Moivre (\(26\) May \(1667\) – \(27\) November \(1754\)). He is also to be credited for his work on the normal distribution and probability theory.

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