# 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.

\(\ds z\) | \(=\) | \(\ds \cos \theta + i \sin \theta\) | ||||||||||||

\(\ds \frac {\d z} {\d \theta}\) | \(=\) | \(\ds -\sin \theta + i \cos \theta\) | Derivative of Sine Function, Derivative of Cosine Function, Linear Combination of Derivatives | |||||||||||

\(\ds \) | \(=\) | \(\ds i^2 \sin \theta + i\cos \theta\) | $i^2 = -1$ | |||||||||||

\(\ds \) | \(=\) | \(\ds i \paren {i \sin \theta + \cos \theta}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds i z\) |

$\Box$

### Step 2

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

\(\ds y\) | \(=\) | \(\ds e^{i\theta}\) | ||||||||||||

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

\(\ds \) | \(=\) | \(\ds i y\) |

$\Box$

### Step 3

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

\(\ds \bigvalueat y {\theta \mathop = 0}\) | \(=\) | \(\ds e^{0 i}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||

\(\ds \bigvalueat z {\theta \mathop = 0}\) | \(=\) | \(\ds \cos 0 + i \sin 0\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 1\) |

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

But a particular solution to a differential equation is unique.

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

$\blacksquare$