The Moivre's formula

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:

Moivre's formula tells us that:

Demonstration

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:

1. Using the argument of a complex number

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,

$$ \forall \theta \in \hspace{0.05em} \mathbb{R}, \enspace n \in \mathbb{Z}, $$
$$ \Bigl[cos(\theta) + isin(\theta)\Bigr]^n = cos(n\theta) + isin(n\theta) \qquad (Moivre ) $$



2. Using the exponential form of a complex number

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,

$$ \forall \theta \in \hspace{0.05em} \mathbb{R}, \enspace n \in \mathbb{Z}, $$
$$ \Bigl[cos(\theta) + isin(\theta)\Bigr]^n = cos(n\theta) + isin(n\theta) \qquad (Moivre ) $$