= Solution
Put $q=|z|^2$ and retain the domain restriction $z\ne0$. Then $z+1/\bar z=(q+1)/\bar z$, so the continued fraction simplifies to
$$
\boxed{\zeta=\frac{q+1}{\bar z(q+2)}
=\frac{z(q+1)}{q(q+2)}.}
$$
The other denominators are nonzero automatically because $q+1,q+2>0$. Writing $z=x+iy$, the real-part condition becomes
$$
\frac{x(q+1)}{q(q+2)}=\frac{x+1/4}{q}.
$$
Multiplying by $q(q+2)$ and rearranging gives $q+4x+2=0$. Thus the first <geometric locus> in the <complex plane> is the <circle>
$$
\boxed{(x+2)^2+y^2=2,\quad\text{or equivalently }|z+2|=\sqrt2.}
$$
Its center is $-2$ and radius is $\sqrt2$; it does not contain the excluded point $z=0$ on its circumference.
The <principal complex logarithm> has imaginary part equal to the <complex argument>, so its equation gives the open ray
$$
z=re^{i\pi/3},\qquad r>0,
\quad\text{equivalently }y=\sqrt3x,\ x>0.
$$
If one interprets the logarithm as multivalued, requiring a value with imaginary part $\pi/3$ gives the same ray. Expanding $z^2$ in the other equation gives $2ixy=2ix$, hence $x(y-1)=0$. Its solution is the union of the entire imaginary axis and the horizontal line $y=1$.
These latter loci bound a <right triangle> with vertices
$$
O=0,\qquad P=i,\qquad Q=\frac1{\sqrt3}+i.
$$
The bounded open region is \b[$0<x<1/\sqrt3$, $\sqrt3x<y<1$]. Its closure is the closed <triangle>. Although the logarithm is undefined at $O$, the imaginary-axis locus includes that vertex. The earlier <circle> is a separate locus and is not a side of this <triangle>.
The right angle is at $P$, so the <circumcenter> is the midpoint of hypotenuse $OQ$:
$$
z_c=\frac1{2\sqrt3}+\frac i2,\qquad R=\frac{|Q|}{2}=\frac1{\sqrt3}.
$$
Thus its <circumcircle> has the complex equation
$$
\boxed{\left|z-\left(\frac1{2\sqrt3}+\frac i2\right)\right|^2=\frac13.}
$$
To find the <incircle>, let its radius be $r$. Tangency to $x=0$ and $y=1$ puts its center at $z_i=r+i(1-r)$. The distance from that point to $y-\sqrt3x=0$ is $(1-r-\sqrt3r)/2$. Requiring this distance to be $r$ gives
$$
r=\frac1{3+\sqrt3}=\frac{3-\sqrt3}{6}.
$$
Therefore
$$
\boxed{|z-[r+i(1-r)]|^2=r^2,\qquad r=\frac{3-\sqrt3}{6}.}
$$
This tangent <circle> is indeed the largest contained <circle>. For any contained disk of radius $\rho$, its center has distances $d_j\geq\rho$ from the three side lines. Splitting the <triangle> into <triangles> with that center as vertex gives area $\Delta=\tfrac12\sum_j\ell_jd_j\geq s\rho$, where $s$ is its semiperimeter. Here $\Delta=1/(2\sqrt3)$ and $s=(1+\sqrt3)/2$, so $\rho\leq\Delta/s=r$. Our <circle> attains this <upper bound>, proving maximality rather than just tangency.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ia/paper-1-complex-loci.png]
{title=Complex loci and the bounded triangle with its circumcircle and incircle}
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