Euler's Formula/Real Domain/Proof 1

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


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


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
\(\ds \) \(=\) \(\ds i^2 \sin \theta + i\cos \theta\) $i^2 = -1$
\(\ds \) \(=\) \(\ds i \paren {i \sin \theta + \cos \theta}\)
\(\ds \) \(=\) \(\ds i z\)


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


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