Why is e^(pi * i) + 1 = 0?

Question

I have seen this equation countless times, but I want to know the proof or at least the gist of where it comes from.

Answer 1

In general e^(i*x)=cos(x)+i*sin(x). Plug in x=pi to get the equation you wanted.

Of course, this just pushes off the result that needs to be demonstrated to the formula for e^(i*x). There are a couple of ways to show this. The first simply plugs i*x in for x in the power series for e^x. The powers of i are regular and when you break the series into even and odd terms, you get the series for cos(x) and sin(x).

Alternatively, realize that e^(ix) is a solution of the differential equation
dy/dx=i*y
and so is cos(x)+i*sin(x). Both also satisfy y(0)=1. By uniqueness results for solutions of differential equations, the two must be equal.

Answer 2

-1 has three cube roots, 1 real and 2 complex.

The roots are:

e^(i*pi)
e^(i*pi/3)
e^(-i*pi/3)

Raising any one of these the the third power yields -1. Thus they are all the solutions to x^3 +1 = 0.

Answer 3

Check out Euler's formula and DeMoivre's formula.

Answer 4

e^(pi * i) + 1
=cospi+isinpi+1
=-1+0+1
=0
because
e^ix=cosx+isinx

Answer 5

you know from DeMoivre's formula.

e^it = cos t + i sin t

put t = pi and kowing cos pi = -1 sin pi =0

e^pi*i = -1
or e^(pi*i) + 1 = 0