# Famous Theorems of Mathematics/Euler's equation

Euler's equation states that,

${\displaystyle \exp i\theta =i\sin \theta +\cos \theta }$

From which, the famous identity,

${\displaystyle \exp i\pi +1=0}$

may be deduced.

## Proof

Definition 1
${\displaystyle \exp \varphi =\sum _{n=0}^{\infty }{\frac {1}{n!}}\varphi ^{n}}$
Definition 2
${\displaystyle \sin \varphi =\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}\varphi ^{2n+1}}$
Definition 3
${\displaystyle \cos \varphi =\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}\varphi ^{2n}}$

The following results will be assumed;

${\displaystyle \sin \pi =0}$
${\displaystyle \cos \pi =-1}$
Theorem 1
${\displaystyle \exp i\theta =\cos \theta +i\sin \theta }$
Proof

By definition 1,

${\displaystyle \exp i\theta =\sum _{n=0}^{\infty }{\frac {1}{n!}}(i\theta )^{n}}$

Observe that one may split the summation into two,

${\displaystyle \exp i\theta =\sum _{n=0}^{\infty }{\frac {1}{(2n)!}}(i\theta )^{2n}+\sum _{n=0}^{\infty }{\frac {1}{(2n+1)!}}(i\theta )^{2n+1}}$

Evaluating,

${\displaystyle \exp i\theta =\sum _{n=0}^{\infty }{\frac {i^{2n}}{(2n)!}}\theta ^{2n}+\sum _{n=0}^{\infty }{\frac {i^{2n+1}}{(2n+1)!}}\theta ^{2n+1}}$
${\displaystyle \exp i\theta =\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}\theta ^{2n}+i\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}\theta ^{2n+1}}$

By definition 2 and 3,

${\displaystyle \exp i\theta =\cos \theta +i\sin \theta }$ ${\displaystyle \blacksquare }$
Theorem 2
${\displaystyle \exp i\pi +1=0}$
Proof

By theorem 1,

${\displaystyle \exp i\pi =\cos \pi +i\sin \pi }$
${\displaystyle \exp i\pi =-1+0i}$

Hence,

${\displaystyle \exp i\pi +1=0}$ ${\displaystyle \blacksquare }$