Prove that e^iω = cos ω + i sin ω......?

Question

where i is the imaginary unit, such that i = √(-1)

Answer 1

Jim Burnell offered an excellent proof of this that doesn't involve infinite series, which was the original method of discovery. I'll repost what he posted some months ago:

Calculus: Start with

z = cos(x) + sin(x) i
and notice that when x = 0, z = 1. Then differentiate,

dz/dx = -sin(x) + cos(x) i
dz/dx = sin(x) i2 + cos(x) i
dz/dx = [cos(x) + sin(x) i]i
dz/dx = zi
(1/z)dz/dx = i
ln(z) = xi + C
for some constant C, by indefinite integration. Now use the fact that when x = 0, z = 1, to conclude that C = 0. Thus

ln(z) = xi
z = e^xi
e^xi = cos(x) + sin(x) i

You can find more of his work via the link given here:

Answer 2

It's an odd concept, isn't it, that you can take any number to an imaginary power? Well, Euler's Formula states how you can.

Here's the proof:

Are you familiar with Taylor series? One of the theories of Taylor series is that it no longer becomes an approximation of a function, but the definition of a function, as you take the amount of terms in a Taylor series out to infinity.

This proof is complex, but here's somethings you need to recall:

Power Identities of i
1. i^0 = 1
2. i^1 = i
3. i^2 = -1
4. i^3 = -i
5. i^4 = 1

If you notice, these values oscillate as you increase them between i, -1, -i, and 1 as you increase the power. This is important.

Taylor Series definition of e^x, cos x, sin x when taken to an infinite number of terms:

e^x = 1 + x + 1/2(x^2) + 1/3(x^3) + 1/4(x^4) .....
cos x = 1 - 1/2(x^2) + 1/4(x^4) - 1/6(x^6) + ....
sin x = x - 1/3(x^3) + 1/5(x^5) - 1/7(x^7) + .....

Therefore:

e^iz = 1 + iz + 1/2(iz)^2 + 1/3(iz)^3 + 1/4(iz)^4 + 1/5(iz)^5 + .....

Now remember how when you took powers of i, it oscillated? Plug those oscillations in:
e^iz = 1 + iz - 1/2(z)^2 - 1/3 i(z)^3 + 1/4(z)^4 + 1/5 i(z)^5 + ...

Reorganize, depending on i, and factor the i out:

e^iz = [ 1 - 1/2(z)^2 1/4(z)^4 + ....] + i [z - 1/3(z)^3 + 1/5(z)^5 + ....]

Those are the Taylor series definitions of sin x and cos x (or z, in this case. But variable names don't matter) So, substitute:

e^iz = cos z + i*sin z

Q.E.D.

Answer 3

There are many ways to prove this to yourself, but probably the most straightforward is to write cos w and sin w as infinite series:

sin w = Sum (i=1 to inf) w^(2i+1)/(2i+1)!

cos w = Sum (i=1 to inf) w^2i/(2i)!

Just form cos w + i sin w out of these, and you will get the infinite series expansion for exp(iw).

Another demonstration is possible with differential equations.

Answer 4

Use Taylor series expansion

http://en.wikipedia.org/wiki/Taylor_series#Examples

Answer 5

:S