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Then we have, $\left|z^2\right|+Re(az)+b=0,$\\  By substitution we get the following: $x^2+y^2 + Re((c+di)(x+yi))+b =0$ \\  $\Leftrightarrow x^2+y^2 By algebra, $x^2+y^2  + Re(cx+cyi+dxi+dyi^2)+b=0$\\ $\Leftrightarrow $  x^2+y^2 + Re(cx-dy+i(dx+cy))+b=0$\\ $\Leftrightarrow x^2 +y^2 + cx-dy + b=0$\\  $\Leftrightarrow (x^2+cx) +(y^2-dy) +b =0$\\  $\Leftrightarrow (x+\frac{1}{2}c)^2 + (y-\frac{1}{2}d)^2 = -b + \frac{1}{4}(c^2+d^2)$ by completing the square\\