Euler’s formula proof

Reading An Introduction to Laplace Transforms and Fourier Series I reach the point where it is stated, rather axiomatically, that: e^{ix} = \cos x + i \sin x .

This is a beautiful formula and has always suggested to me some sort of mystical inner mathematical harmony (yes, I am a materialist, but I cannot help it).

But these days I also want to see the proof, so here is one:

We know that complex numbers can be described in polar co-ordinates:

z = |z| (\cos \theta + i\sin \theta)

So too e^{ix} = r(\cos \psi + i \sin \psi) where r and \psi depend on x .

Now (and applying the product rule) \frac{d}{dx}e^{ix} = ie^{ix} = \frac{dr}{dx}(\cos \psi + i\sin \psi) + \frac{d \psi}{dx}r(-\sin \psi + i \cos \psi)

So we equate the real and imaginary sides of both sides of this equality we have:

ie^{ix} = (\cos \psi \frac{dr}{dx} - r \sin \psi \frac{d \psi}{dx}) + i(\sin \psi \frac{dr}{dx} + r \cos \psi \frac{d\psi}{dx})

Then, recalling e^{ix} = r(\cos \psi + i \sin \psi) , we have ir (\cos \psi + i \sin \psi) = ir \cos \psi - r sin \psi = ( \cos \psi \frac{dr}{dx} - r \sin \psi \frac{d\psi}{dx}) + i( \sin \psi \frac{dr}{dx} + r \cos \psi \frac{d\psi}{dx})

By inspection we can see that \frac{dr}{dx} = 0 and \frac{d\psi}{dx} = 1, giving us:

ie^{ix} = - r \sin \psi + ir \cos \psi and multiplying both sides by i we have: -e^{ix} = -r \cos \psi - i r \sin \psi

Reversing the signs: e^{ix} = r \cos \psi + i r \sin \psi

But what of r and \psi ? Well, we have \frac{dr}{dx} = 0 and \frac{d \psi}{dx} = 1.

So r is constant with respect to x while \psi varies as x .

If we set x = 0 then e^{i \times 0} = 1 – a wholly real number, so \sin \psi = 0 and \psi = 0. Thus r ( \cos 0 ) = e^0 = r = 1 and we can replace \psi with x throughout.

Hence: e^{ix} = \cos x + i \sin x .

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