= Solution
Write $z=x+iy$. Dividing by two gives
$$
\sqrt{(x-a)^2+y^2}=x+a.
$$
The left side is nonnegative, so squaring is reversible only when $x+a\ge0$. Squaring yields $y^2=4ax$.
If $a>0$, this equation implies $x\ge0$, so the sign condition holds automatically. \b[The locus is the <parabola>]
$$
\boxed{y^2=4ax,\qquad a>0,}
$$
with vertex at the origin, <focus of a conic> $(a,0)$ and <directrix of a conic> $x=-a$. Indeed the original equation equates distance from the <focus of a conic> with distance from the <directrix of a conic> on its right side.
If $a<0$, the squared equation forces $x\le0$, making $x+a<0$. This violates the sign condition. Thus \b[the locus is empty for $a<0$]; squaring without checking that condition would introduce a spurious <parabola>.
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