TITLE: How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
QUESTION [173 upvotes]: Could you provide a proof of Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?

REPLY [3 votes]: Let $y=\cos \phi+i\sin \phi$ $...(1)$
Differentiating both sides of equation (1) with respect to $\phi$, we get, 
$\frac{dy}{d\phi}=-\sin \phi+i\cos \phi$
$\implies \frac{dy}{d\phi}=i(\cos \phi-\frac{1}{i}\sin \phi)$
$\implies \frac{dy}{d\phi}=i(\cos\phi+i\sin \phi)$
$\implies \frac{dy}{d\phi}=iy$
$\implies\frac{1}{y}dy=id\phi$ $...(2)$
Integrating both sides of equation (2), we get,
$\int\frac{1}{y}dy=\int id\phi$
$\implies \ln(y)=i\phi+c$ $...(3)$
Substituting $\phi=0$ in equation (1), we get, 
$y=\cos 0+i\sin 0$
$\implies y=1$
Substituting $\phi=0$ and $y=1$ in equation (3) we get,
$\ln(1)=c$
$\implies c=0$
Substituting $c=0$ in eqaution (3) we get, 
$\ln(y)=i\phi$
$e^{i\phi}=y$
$\therefore e^{i\phi}=\cos \phi+i\sin \phi$