Euler's Formula/Real Domain/Proof 1

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Theorem

Let $\theta \in \R$ be a real number.

Then:

$e^{i \theta} = \cos \theta + i \sin \theta$


Proof

Consider the differential equation:

$D_z \map f z = i \cdot \map f z$


Step 1

We will prove that $z = \cos \theta + i \sin \theta$ is a solution.

\(\displaystyle z\) \(=\) \(\displaystyle \cos \theta + i \sin \theta\)
\(\displaystyle \frac {\d z} {\d \theta}\) \(=\) \(\displaystyle -\sin \theta + i \cos \theta\) Derivative of Sine Function, Derivative of Cosine Function, Linear Combination of Derivatives
\(\displaystyle \) \(=\) \(\displaystyle i^2 \sin \theta + i\cos \theta\) $i^2 = -1$
\(\displaystyle \) \(=\) \(\displaystyle i \paren {i \sin \theta + \cos \theta}\)
\(\displaystyle \) \(=\) \(\displaystyle i z\)

$\Box$


Step 2

We will prove that $y = e^{i\theta}$ is a solution.

\(\displaystyle y\) \(=\) \(\displaystyle e^{i\theta}\)
\(\displaystyle \frac {\d y} {\d \theta}\) \(=\) \(\displaystyle i e^{i \theta}\) Derivative of Exponential Function, Chain Rule, Linear Combination of Derivatives
\(\displaystyle \) \(=\) \(\displaystyle i y\)

$\Box$


Step 3

Consider the initial condition $\map f 0 = 1$.

\(\displaystyle \bigintlimits y {\theta \mathop = 0} {}\) \(=\) \(\displaystyle e^{0 i}\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)
\(\displaystyle \bigintlimits z {\theta \mathop = 0} {}\) \(=\) \(\displaystyle \cos 0 + i \sin 0\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)

So $y$ and $z$ are both particular solutions.

But a particular solution to a differential equation is unique.


Therefore $y = z$, that is, $e^{i \theta} = \cos \theta + i \sin \theta$.

$\blacksquare$