⁴The proof of this is remarkably simple if we take the Taylor expansions of an exponential function and of a sine and cosine: $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$, $\mathit{cos}(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$ and $\mathit{sin}(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$. If we remember that $j^0=1$, $j^1=j$, $j^2=-1$ and $j^3=-j$ then it immediately follows that $e^{\mathit{jx}}=\mathit{cos}(x)+j\cdot \mathit{sin}(x)$.