## Algebra 2 (1st Edition)

$e^{\pi i}+1=0$
We are given that $e^{a+ib}=e^a (\cos b+i \sin b)$ Plug $a=0; b=\pi$ $e^{0+\pi i}=e^{0} (\cos \pi+i \sin \pi)$ This gives: $e^{\pi i}=-1+0$ and $e^{\pi i}+1=0$ Hence, the result has been proved.