This answer should go as a comment at George's analysis, but I had an error in the comments and also I couldn't format them properly. So here it goes.
To force a symmetry around the imaginary axis means to introduce a definition independent of the ansatz by the formal powerseries. We'll have to check, whether they finally match.
Assume some circle of radius $c$ over the origin with center at $0+c*î$
Then we choose $z_0 = 2*c*î$, just the top of the circle.
Using the functional equation $z_1 = z_0^2+z_{-1} $ together with the symmetry-assumtion and the assumtion, that $z_1$,$z_0$,$z_{-1}$ are on the circumference of the circle we can uniquely determine all needed coordinates and have thus a "germ" for the iteration. What we get is the following.
denote $s= \sqrt{1-4*c^2} $
Then we get
$ z_{-1} = 2 c^2 + i*c*(1+s) $
$ z_0=i*2*c $
$ z_1 = - 2 c^2+i*c*(1+s) $
and the list of the trajectory counterclockwise through the second quadrant
$ z_0 = i* 2*c $
$ z_1 = -2*c^2 + i*(s + 1)*c $
$ z_2 = 4*c^4 - (s+1)^2*c^2 + i*(-4*(s+1)*c^3 + 2*c) $
$ z_3 = 16*c^8 - 24*(s+1)^2*c^6 + (s^4 + 4*s^3 + 6*s^2 + 20*s + 17)*c^4 - 6*c^2 $
$ + i*((-32*s - 32)*c^7 + (8*s^3 + 24*s^2 + 24*s + 24)*c^5 + (-4*s^2 - 8*s - 4)*c^3 + (s + 1)*c) $
(The output can be much more simplified)
The resulting trajectory is very near to that which was computed using the truncated powerseries. It is symmetric by construction, however it leaves the circumference of the circle already at $z_2$
[update] here is a comparision of the "circularity" of the trajectories (as it was already computed) using the powerseries and that using the above ansatz with the assumtion of a symmetric and circular initializing around the center $(0,i*c)$
[end update]
