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\begin {problem}[1.10]\\  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}  \textit{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 $x,y,c,d \in \mathbb{R}$.\\  So, assume $\left|z^2\right|+Re(az)+b=0$