= Solution
{c}
Write $z=x+iy$, with $y\ne0$ because the <triangle> is nondegenerate. Let $w=u+iv$ be its <circumcenter>. Equality of the <complex modulus> values $|w|=|w-1|$ gives $u=1/2$. Equality of $|w|=|w-z|$, after expanding the squares, gives
$$
2\operatorname{Re}(w\bar z)=|z|^2,\qquad x+2vy=x^2+y^2.
$$
Thus the <real part> and <imaginary part> determine
$$
w=\frac12+i\frac{x^2+y^2-x}{2y}.
$$
Using the <complex conjugate> $\bar z=x-iy$, we also have
$$
\frac{z(1-\bar z)}{z-\bar z}
=\frac{x-x^2-y^2+iy}{2iy}
=\frac12+i\frac{x^2+y^2-x}{2y}.
$$
Therefore
$$
\boxed{w=\frac{z(1-\bar z)}{z-\bar z}.}
$$
The nonzero <denominator> is exactly the nondegeneracy condition. If $z$ is real, the three distinct collinear <points> do not have a finite <circumcenter>.
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