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An equation $|az+b\overline{z}|=ux+v$, with $z=x+iy$ and real coefficients, becomes $(a+b{)}^2{x}^2+(a-b{)}^2{y}^2=(ux+v{)}^2$ after squaring. If $(a+b{)}^2-u^2=(a-b{)}^2>0$, completing the square gives a circle. The original locus is only the part where $ux+v\ge 0$. Checking this sign on the candidate circle prevents extraneous points introduced by squaring. This method combines the complex modulus with elementary conic geometry.

The complex modulus of $z=x+iy$ is $|z|=\sqrt{x^2+y^2}=\sqrt{z\overline{z}}$. It is the Euclidean norm of $(x,y)$ in the complex plane, and satisfies $|zw|=|z||w|$ and the triangle inequality.

Write $z=x+iy$. Using the complex conjugate, the squared complex modulus on the left is $(\alpha +\beta {)}^2{x}^2+(\alpha -\beta {)}^2{y}^2$. Squaring the equality and dividing by $(\alpha -\beta {)}^2$, which is nonzero, givesCompleting the square identifies the candidate circle:Its 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, $x\ge 2\sqrt{\alpha \beta }-(\alpha +\beta )$, so the original right-hand side satisfiesThus 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.