This question has been around for a while and has some decent answers. But I want to suggest a simpler and more intuitive version. (And it will be easier to follow if I replace the variable [math]x[\/math] by [math]t[\/math] so as not to confuse it with the real part of the complex function value.)<\/p>\n
First, to define [math]f(z)=e^z[\/math] without using trigonometry or ever mentioning trig functions, we can use either the power series or the complex differential equation [math]f\u2019=f[\/math] with [math]f(0)=1[\/math]. And either way we get [math]\\frac{d}{dt}e^{it}=ie^{it}[\/math].<\/p>\n
Now multiplication by [math]i[\/math] just rotates the complex plane by a right angle, so the curve in the plane given parametrically by [math](x(t),y(t))[\/math] with [math]x(t)+iy(t)=e^{it}[\/math] has a tangential velocity vector which is always perpendicular to its position vector and equal in magnitude.<\/p>\n
Since it starts at [math]t=0[\/math] at [math](x,y)=(1,0)[\/math] the curve is just the unit circle centred at the origin.<\/p>\n
And since its velocity vector is always of length 1, if we think of the parameter [math]t[\/math] as representing time, then the point moves with speed 1 and so the time taken to complete a circuit, ie the period of [math]e^{it}[\/math], is just the same as the circumference of the unit circle (commonly denoted by [math]2\\pi[\/math] ).<\/p>\n