Prove Euler's Identity

Question

I have always wondered how to prove Euler's identity (e^i*theta)=cos(theta)+i*sin(theta). The whole thing just seems so glorious.

Answer 1

A simple way to prove this is to find the second derivative of each side of the identity separately:

Let x=e^iθ, note that x(0)=1

dx/dθ=ie^iθ, note that x'(0)=i

d²x/dθ²=-e^iθ=-x

Let y=cosθ+isinθ, note that y(0)=1

dy/dθ=-sinθ+icosθ, note that y'(0)=i

d²y/dθ²=-cosθ-isinθ=-y

This shows that the second derivatives are identical in terms of their primitives, that is, x=y, and that substitution of θ=0 in the primitive and first derivatives yield the same result. Therefore x≡y.

Another way is to use the Taylor expansion of each series:

e^iθ≡1+iθ+(iθ)²/2!+...+∑(iθ)ⁿ/n!+...=

1-θ²/2!+θ⁴/4!-... (series for cosθ) + i(θ-θ³/3!+θ⁵/5!-...) (i × series for sinθ)=

cosθ+isinθ