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Fix $a \in \mathbb{C}$ and $b \in \mathbb{R}.$ Show that the equation $\left|z^2\right|+Re(az)+b=0$ has a solution if and only if $\left|a^2 \right| \geq 4b.$ When solutions exist, show the solution set is a circle.  \begin{proof}  Let $a \in \mathbb{C}$, $b \in \mathbb{R}$ and $z \in \mathbb{C}$. Then $z=x+yi$ and $a=c+di$ for some $x,y,c,d \in \mathbb{R}$.\\  Then we have, $\left|z^2\right|+Re(az)+b=0,$\\ $$\left|z^2\right|+Re(az)+b=0,$$\\  By substitution we get the following: $x^2+y^2 $$x^2+y^2  + Re((c+di)(x+yi))+b =0$ =0$$  \\ By algebra, $$x^2+y^2 + Re(cx+cyi+dxi+dyi^2)=-b$$\\  $$ x^2+y^2 + Re(cx-dy+i(dx+cy))=-b$$\\  $$ x^2 +y^2 + cx-dy = -b$$\\