Why is e^(pi.i) equal to -1?

[sebasalekhine7]

ok, here it goes, why is e^(pi.i)=-1 ?

 

[Integral]

It comes form Eulers relationship

$$e^{ix} = \cos(x) + i \sin(x)$$

Edit: LOL, I fixed it already!
Now I am moving this to Math.

 

[freemind]

Any complex number $$a + bi$$ can be written in the form $$r(\cos{\theta} + i\sin{\theta})$$. De Moivre's theorem states that this in turn can be written in the form $$re^{i\theta}$$. That is, $$re^{i\theta} = r(\cos{\theta} + i\sin{\theta})$$. Plug in $$\theta = \pi$$ and behold the magic!

 

[sebasalekhine7]

Was it Euler? well, in that case u would have to use Taylor's expansion series and yes, it would work. Why is this important at all then?

 

[freemind]

It is commonly regarded to be one of the most (ahem) beautiful and elegant mathematical relationships in our universe (yes, our ). C'mon, wouldn't you agree that it is beautiful, succintly relating the 5 most important numbers in mathematics?? ($$e^{\pi i} + 1 = 0$$)

 

[Integral]

It is commonly regarded to be one of the most (ahem) beautiful and elegant mathematical relationships in our universe (yes, our ). C'mon, wouldn't you agree that it is beautiful, succinctly relating the 5 most important numbers in mathematics?? ($$e^{\pi i} + 1 = 0$$)

In addition it uses each of the fundamental mathematical operations, addition, multiplication, exponentiation, and equality. It is consider mathematical poetry.