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:

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

Step 1

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

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

$\Box$

Step 2

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

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

$\Box$

Step 3

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

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

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

But a specific solution to a differential equation is unique.

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

$\blacksquare$