= Solution
Write $z=x+iy$. Using the <complex conjugate>, the squared <complex modulus> on the left is $(\alpha+\beta)^2x^2+(\alpha-\beta)^2y^2$. Squaring the equality and dividing by $(\alpha-\beta)^2$, which is nonzero, gives
$$
x^2+y^2-4\sqrt{\alpha\beta}\,x-(\alpha-\beta)^2=0.
$$
Completing the square identifies the candidate <circle>:
$$
\boxed{(x-2\sqrt{\alpha\beta})^2+y^2=(\alpha+\beta)^2.}
$$
Its \b[centre is $2\sqrt{\alpha\beta}$ on the real axis and its radius is $\alpha+\beta$] in the <complex plane>.
It remains to exclude extraneous points introduced by squaring. On this <circle>, $x\geq 2\sqrt{\alpha\beta}-(\alpha+\beta)$, so the original right-hand side satisfies
$$
2\sqrt{\alpha\beta}\,x+(\alpha-\beta)^2\geq(\alpha+\beta)(\alpha+\beta-2\sqrt{\alpha\beta})>0.
$$
Thus taking the nonnegative square root recovers the original equality at every point of the <circle>. This is a <circle from an affine complex-modulus equation>, with the sign check essential to the geometric conclusion.
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