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\begin {problem}[1.10]\\ {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} \end{proof}  \end{problem}