I think I can summarize Needham's argument without having to draw any pictures. But it's a lot better with the pictures. Imagine a particle beginning at (1,0) in the plane, a.k.a. at 1 in the complex plane. This particle moves such that its velocity is always perpendicular to the vector from the origin to its current position, and its speed is always equal to the magnitude of its distance from the origin. That is, with s(t) being its position, s'(t) = i * s(t). Thus s(t) = e^(it), up to some constants which turn out to be 1 because of the given initial conditions. But velocity perpendicular to displacement vector means circular motion: the velocity never has a radial component so the distance from the origin never changes. Furthermore, the acceleration is perpendicular to the velocity here, and thus the magnitude of the velocity never changes; it's always 1 as well. Thus the particle is in uniform circular motion around the origin, at speed 1 (distance unit per second, hence also radian per second) so s(t) = cos(t) + i sin(t). --Joshua