Let \( n \in \mathbb{Z}\) be an integer, and \( z \in \mathbb{C}\) be a complex number having as a module \( |z|= 1\) under its trigonometric form, such as:
$$ z = cos(\theta) + isin(\theta) $$
Moivre's formula tells us that:
$$ \forall \theta \in \hspace{0.05em} \mathbb{R}, \enspace n \in \mathbb{Z}, $$
$$ \Bigl[cos(\theta) + isin(\theta)\Bigr]^n = cos(n\theta) + isin(n\theta) \qquad (Moivre ) $$
Let \( n \in \mathbb{Z}\) be an integer, and \( z \in \mathbb{C}\) be a complex number having as a module \( |z|= 1\) under its trigonometric form, such as:
$$ z = cos(\theta) + isin(\theta) $$
Raising both sides to the power of \(n\), we do have:
$$ z^n = \Bigl[cos(\theta) + isin(\theta)\Bigr]^n $$
We know from the property of the argument of a complex number raised to an integer power that:
$$ \forall z \in \mathbb{C}, \enspace \forall n \in \mathbb{Z}, \enspace arg(z^n) = n . arg(z) $$
So in our case that,
$$ arg(z^n) = n . arg(z) $$
$$ arg(z^n) = n \theta $$
If the argument of the complex number \( z^n \) is \( n \theta \), then the latter can we written as:
$$ z^n = cos(n\theta) + isin(n\theta) $$
And finally,
$$ \forall \theta \in \hspace{0.05em} \mathbb{R}, \enspace n \in \mathbb{Z}, $$
$$ \Bigl[cos(\theta) + isin(\theta)\Bigr]^n = cos(n\theta) + isin(n\theta) \qquad (Moivre ) $$
Using the exponential form of a complex number, we can directly notice that:
$$ cos(\theta) + isin(\theta) = e^{i\theta} $$
Now, raising both sides to the power of \(n\):
$$ \Bigl[cos(\theta) + isin(\theta)\Bigr]^n = \bigl(e^{i\theta}\bigr)^n $$
$$ \Bigl[cos(\theta) + isin(\theta)\Bigr]^n = e^{in\theta} $$
And finally,
$$ \forall \theta \in \hspace{0.05em} \mathbb{R}, \enspace n \in \mathbb{Z}, $$
$$ \Bigl[cos(\theta) + isin(\theta)\Bigr]^n = cos(n\theta) + isin(n\theta) \qquad (Moivre ) $$