Euler's Formula/Real Domain/Proof 1

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.

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

$\blacksquare$