How to prove e^ix=cos x + i sin x

[HeilPhysicsPhysics]

How to prove e^ix=cos x + i sin x

 

[d_leet]

How to prove e^ix=cos x + i sin x

One way is to start with the taylor series for e^x and then change x to ix and remembering that
i² = -1, i³ = -i, and i⁴ = 1 you can rearrange the series and show that this is equal to the other side of that equation.

 

[Data]

The proof depends on your definitions (for example, if you define \cos{x} as (e^{ix}+e^{-ix})/2 and \sin{x} as (e^{ix} - e^{-ix})/(2i) then it's pretty easy!).

It's also clear if you start from the Taylor series of the three functions about the origin as d_leet suggested.

You can also start from the differential equation for e^x,

y^\prime = y, y(0)=1.

Write

e^{ix} = f(x) + ig(x),

(where f, g are real-valued)

so that f(0) = 1 and g(0) = 0

then you must have (by the chain rule)

ie^{ix} = f^\prime(x) + ig^\prime(x),

or, rearranging,

e^{ix} = \frac{1}{i}f^\prime(x) + g^\prime(x) = g^\prime(x) - if^\prime(x).

Since e^0 = 1 that means g^\prime(0) = 1 and f^\prime(0) = 0. Now differentiate again and you get

ie^{ix} = g^{\prime \prime}(x) - if^{\prime \prime}(x)

or

e^{ix} = -f^{\prime \prime}(x) - ig^{\prime \prime}(x).

Comparing the real and imaginary parts with our original e^{ix} = f(x) + ig(x) you get two differential equations with initial conditions:

f^{\prime \prime}(x) = -f(x), f^{\prime}(0) = 0, f(0) = 1

and

g^{\prime \prime} (x) = -g(x), g^{\prime}(0) = 1, g(0) = 0.

It's easy to check that f(x) = cos(x) and g(x)= sin(x) satisfy these DEs, and by a theorem in the theory of DEs the solutions are unique.

 

[Edgardo]

Let z = \rm{cos}\theta + i \cdot \rm{sin}\theta

Then derive z:

\frac{\rm{d}z}{\rm{d}\theta} = - \rm{sin}\theta + i \cdot \rm{cos}\theta = -1 (\rm{sin}\theta - i \cdot \rm{cos}\theta) = i^2 (\rm{sin}\theta - i \cdot\rm{cos}\theta ) = i (i \cdot \rm{sin}\theta - i^2 \cdot \rm{cos}\theta) = i(\rm{cos}\theta + i \cdot \rm{sin}\theta) = i \cdot z

This is a differential equation:
\frac{\rm{d}z}{\rm{d}\theta} = i z
whose solution is

z = e^{i \theta}

Thus, \rm{cos}\theta + i \rm{sin}\theta = e^{i \theta}

 

[eljose]

Umm.. in fact i think that the "historical" development fo the formula came with euler when studying the "Harmonic (classical) oscilator"..

y'' + \omega ^2 y =0 trying y=exp(ax) then you get..

a^2 +\omega ^2 =0 so + \sqrt (-1) \omega =a (and the same

with a minus sign)

-On the other hand the long-known solution for the Harmonic movement was y= Asin (\omega t) +B Cos(\omega t) then if you set the initial condition y(0)=0 you get the identity for sin. and using the well-known formula cos^2 (x) + sen^2 (x) =1 oh..if today math were so easier...i would have became famous years ago...