Euler's Formula/Real Domain/Proof 1
Jump to navigation
Jump to search
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.
This article needs to be linked to other articles. In particular: "Specific solution" and the proof that such a specific solution is unique. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
Therefore $y = z$, that is, $e^{i \theta} = \cos \theta + i \sin \theta$.
$\blacksquare$