We first establish by mathematical induction the statement \(P_n\):
\[z^n = r^n[\cos(n\theta) + i\sin(n\theta)], n \in \mathbb{Z}^+, n \ge 0.\]
In this proof, we demonstrate the power of the principle of mathematical induction from Unit \(1.15\).
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&nbsp;\(P_{k+1}\) 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
\(\begin{aligned} &= z^k \cdot z = r^k[\cos(k\theta) + i\sin(k\theta)] \cdot r[\cos\theta + i\sin\theta] \\ &= r^{k+1}[\cos(k\theta + \theta) + i\sin(k\theta + \theta)] \\ &= r^{k+1}[\cos((k+1)\theta) + i\sin((k+1)\theta)], \end{aligned}\)
and hence&nbsp;\(P_{k+1}\) is true.
Since&nbsp;\(P_0\) is true and \(P_k \Rightarrow P_{k+1}\) for each \(k \in \mathbb{Z}^{+}\), \(k \ge 0\), by Mathematical Induction, it follows that&nbsp;\(P_n\) holds for all integers \(n \ge 0\).
We now turn to the case when&nbsp;\(n\) is a negative integer. It follows immediately that&nbsp;\(m = -n\) is a positive integer.
So, by the result of reciprocation and by \(P_m\), we have
\[z^n = \frac{1}{z^m} = \frac{1}{r^m}[\cos(m\theta) - i\sin(m\theta)] = r^n[\cos(-n\theta) - i\sin(-n\theta)] = r^n[\cos(n\theta) + i\sin(n\theta)].\]
Thus, the result also holds for negative integers \(n\).
All in all, we have that if&nbsp;\(z = r(\cos\theta + i\sin\theta)\) then
\[z^n = r^n[\cos(n\theta) + i\sin(n\theta)]\]
for all \(n \in \mathbb{Z}\).
In particular, when&nbsp;\(r = 1\), 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.

We first establish by **mathematical induction** the statement \(P_n\):

\[z^n = r^n[\cos(n\theta) + i\sin(n\theta)], n \in \mathbb{Z}^+, n \ge 0.\]

In this proof, we demonstrate the power of the principle of mathematical induction from Unit \(1.15\).

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

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

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

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

Since \(P_0\) is true and \(P_k \Rightarrow P_{k+1}\) for each \(k \in \mathbb{Z}^{+}\), \(k \ge 0\), by Mathematical Induction, it follows that \(P_n\) holds for all integers \(n \ge 0\).

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

\[z^n = \frac{1}{z^m} = \frac{1}{r^m}[\cos(m\theta) - i\sin(m\theta)] = r^n[\cos(-n\theta) - i\sin(-n\theta)] = r^n[\cos(n\theta) + i\sin(n\theta)].\]

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

All in all, we have that if \(z = r(\cos\theta + i\sin\theta)\) then

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

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.

Proof of De Moivre's Theorem by induction

Proof of De Moivre's Theorem by induction