The Moivre's formula

Proofs

Let \( n \in \mathbb{Z}\) be an integer, and \( z \in \mathbb{C}\) be a complex number having as a modulus \( |z|= 1\) 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:

   $$ z^n = \Bigl[\cos(\theta) + i \hspace{0.2em} \sin(\theta)\Bigr]^n $$

   We know from the property of the argument of a complex number raised to an integer power that:

   $$ \forall z \in \mathbb{C}, \enspace \forall n \in \mathbb{Z}, \enspace \arg(z^n) = n \cdot \arg(z) $$

   So in our case that,

   $$ \arg(z^n) = n \cdot \arg(z) $$
   $$ \arg(z^n) = n \theta $$

   If the argument of the complex number \( z^n \) is \( n \theta \), then the latter can we written as:

   $$ z^n = \cos(n\theta) + i \hspace{0.2em} \sin(n\theta) $$

   And finally,

   $$ \forall \theta \in \hspace{0.04em} \mathbb{R}, \enspace \forall n \in \mathbb{Z}, $$
   $$ \Bigl[\cos(\theta) + i \hspace{0.2em} \sin(\theta)\Bigr]^n = \cos(n\theta) + i \hspace{0.2em} \sin(n\theta) \qquad \text{(Moivre's formula)}$$

2. Using the exponential form of a complex number

   Using the exponential form of a complex number , we can directly notice that:

   $$ \cos(\theta) + i \hspace{0.2em} \sin(\theta) = e^{i\theta} $$

   Now, raising both sides to the power of \(n\):

   $$ \Bigl[\cos(\theta) + i \hspace{0.2em} \sin(\theta)\Bigr]^n = \bigl(e^{i\theta}\bigr)^n $$
   $$ \Bigl[\cos(\theta) + i \hspace{0.2em} \sin(\theta)\Bigr]^n = e^{in\theta} $$

   And finally,

   $$ \forall \theta \in \hspace{0.04em} \mathbb{R}, \enspace \forall n \in \mathbb{Z}, $$
   $$ \Bigl[\cos(\theta) + i \hspace{0.2em} \sin(\theta)\Bigr]^n = \cos(n\theta) + i \hspace{0.2em} \sin(n\theta) \qquad \text{(Moivre's formula)}$$