If we divide both sides by z, we get
and so integrating both sides with respect to x gives
Substituting into the equation gives .
And since we know is a circle of radius 1, so must be .
Personally I prefer the proof by Maclaurin series.
You can also work with . It's not too hard to show that the modulus is equal to 1 and that the argument is equal to x (I leave this for others who are interested).