Eulers identity

$e^{i}\pi +1=0$

The identity is a special case of Euler's formula from complex analysis, which states that

$e^{ix}=\cos x+i\sin x\,\!$

for any real number x. (Note that the arguments to the trigonometric functions sin and cos are taken to be in radians.) In particular,

$e^{i\pi }=\cos \pi +i\sin \pi .\,\!$

Since

$\cos \pi =-1\,\!$

and

$\sin \pi =0,\,\!$

it follows that

$e^{i\pi }=-1,\,\!$

which gives the identity

$e^{i\pi }+1=0.\,\!$