Proof of DeMoivre's Theorem

We want to prove

e^ib = cos(b) + i sin (b)

You may want to be very sure you believe each step, since this is a proof which makes you think, "yes, yes, yes, how'd you do that?" You might call it proof by surprise.

Define the right-hand-side to be f(b) = cos(b) + i sin(b), so that what we are trying to prove is that f(b) = e^ib. Now

df(b)/db = - sin(b) + i cos(b)
= i [cos(b) + i sin(b)]
= i f(b)
[1/f(b)] df/db = i

Next integrate back, using standard formulae:

Integral {[1/f(b)] df/db db} = Integral {i db}

Integral {1/f df} = Integral {i db}

ln f = i b + ln C

where ln C is just an arbitrary constant.

ln f/C = i b

f/C = e^ib

f = C e^ib

From the original definition of f(b), f(0)=1, so C=1 and

f(b) = e^{i b}

which is exactly what we set out to prove.