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:

$D_z f\left({z}\right) = i \cdot f\left({z}\right)$


Step 1

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

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

$\Box$


Step 2

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

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

$\Box$


Step 3

Consider the initial condition $f \left({0}\right) = 1$.

\(\displaystyle \left.{y}\right \vert_{\theta \mathop = 0}\) \(=\) \(\displaystyle e^{0i}\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)
\(\displaystyle \left.{z}\right \vert_{\theta \mathop = 0}\) \(=\) \(\displaystyle \cos 0 + i \sin 0\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)

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

But a specific solution to a differential equation is unique.


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

$\blacksquare$