# Mathematical Proof

#### f95toli

##### Golden Member
exp(i*pi)=-1 ?
which follows from exp(ix)=cosx+i sinx (Euler's formula)

So are you asking for a proof of Euler's formula?

#### eigen

##### Diamond Member
Originally posted by: f95toli
exp(i*pi)=-1 ?
which follows from exp(ix)=cosx+i sinx (Euler's formula)

So are you asking for a proof of Euler's formula?

Yup, now just look at the taylor series expansions of both terms...prolly a more elegant way.

#### sao123

##### Lifer
I was just "told" that it is true, but i needed someone to show me the calculation.

#### itachi

##### Senior member
err.. he sorta did...
e^ix = cos x + i sin x.. x = pi -> e^ipi = cos pi + i sin pi = -1
ln (e^i*pi) = ln (-1) = i*pi

#### eLiu

##### Diamond Member
taylor expand sin & cos and you're done. I think that's pretty much the most popular method.
Edit: this works b/c the series for sin & cos are uniformly convergent...not too difficult to prove.
You might also benefit from this: sin(z)=[exp(iz)-exp(-iz)]/2i, cos(z)=[exp(iz)+exp(-iz)]/2.

Another way I've seen is to define a complex # z=cos(t)+i*sin(t)

dz/dt = -sin(t) + i*cos(t)...now remember i^2 = -1, so we have i(i*sin(t) + cos(t)) = i*z

Now integrate. The integrating constant is clearly 0...then exponentiate and you're done.