Proof for Euler's Identity using ODEs

Let

z = \cos θ + i \sin θ

Differentiate with respect to $θ$ (Imaginary numbers are just another axis)

\begin{aligned} \frac{d z}{d θ} & = - \sin θ + i \cos θ \\ = i (i \sin θ + \cos θ) \\ = i z \end{aligned}

\begin{aligned} \frac{1}{z} d z & = i d θ \\ \int \frac{1}{z} d z & = \int i d θ \\ \ln z & = i θ + C \\ e^{i θ + C} & = z \end{aligned}

letting $θ = 0$

\begin{aligned} e^{i 0 + C} & = \cos 0 + i \sin 0 \\ e^{0} e^{C} & = 1 \\ C & = 0 \end{aligned}

Thus

e^{i θ} = z = \cos θ + i \sin θ