We wish to show that $e^{iπ}+1=0$. Let $y = \cos θ + i \sin θ$. Differentiating, ${dy} / {dθ} = - \sin θ + i \cos θ = i(i\sin θ + \cos θ)=iy$; therefore, ${dy}/y = i dθ$. Integrating, $\ln y = iθ+C$; thus, $y=e^{iθ}e^C$. Let $θ=0$, so $y(0) = \cos 0 + i \sin 0 = 1$; therefore, $1=e^{i0}e^C=e^C$, $C=0$, and $y=e^{iθ}$. Observe that $y(π)=\cos π + i \sin π=-1$, so $e^{iπ}=-1$, and $e^{iπ}+1=0$. Q.E.D.
From Math with Bad Drawings.
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