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

Question

ive always wanted to know....

Answer 1

e^(iθ) = cos(θ) - isin(θ)

i is the 'complex number'

When θ = π you have:

cos(π) = -1
sin(π) = 0

So you end up with:

e^(iπ) + 1 = cos(π) - isin(π) + 1
==> 1 - 1 = 0

:)

Answer 2

Think of a vector of unit length with its "tail" hooked to the origin of the Complex Plane (vertical axis is the imaginary axis, horizontal axis is the real axis).

You can represent this vector as exp^, where in your example is PI. In effect the number is just the angle of rotation of the vector about the origin of the Complex Plane.

As the argument of the exponential (which is just the power to which the exponential is raised) changes, that vector basically rotates around the origin of the Complex Plane. If the argument is zero, the vector points to the right, along the positive real axis: exp(0) is 1, right?

If the argument of the exponential is i*PI/2, the vector points straight up the positive imaginary axis: exp(i*PI/2). Recall that PI/2 radians is the same as 90 degrees. See how it's working?

So if you let the vector rotate around so that it points along the *negative* real axis, the argument of the exponential is (i*PI), and indeed PI radians is 180 degress.

So a unit vector pointing along the negative real axis can be written as exp(i*PI). Remember, it's a unit vector, so its length is one.

So exp(i*PI) = -1, which is just another way of saying that exp(i*PI) + 1 = 0.

I hope that helps.

Answer 3

It's from Euler's equation, which is
Cos(x) + i Sin(x) = e^(x*i)
So, if x = pi, we have
-1 + 0 = e*(pi*i)

Euler's equation comes from adding the infinite series for Cos(x) and i Sin(x), which equals the infinite series for e^(x*i). That's how it was originally derived. There's a method for determining an infinite series for an arbitrary function f(x), called Taylor's series, which makes use of differentiation.

The interesting question is whether or not Euler's formula can be proven without resorting to infinite series of functions. So, that will be my next question.

Addendum: JimBurnell has given an excellent response to this. See link to another Yahoo! Answers Q&A.

Answer 4

well e^(pi*i)=-1 So the equation is really -1+1=0. Now you know.

Answer 5

The Euler equation is really cool. I was taught it in first year college wave mechanics. It kind of creeps up on you then bang, you get to the most improbable combination of fundamental symbols.

Here is a useful link

Answer 6

e^ix=cos x +i*sin x
e^i*π=cos π+i*sin π =-1+0i
-1+1=0 so
e^(iπ)+1=0