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\end{problem}  \begin{proof}\textit{Proof:} Let $a\in\mathbb{C}$,$b\in\mathbb{R}$, and $z\in\mathbb{C}$.\\  Then $a=d+ei$ and $z=f+gi$ for some $d,e,f,g\in\mathbb{R}$.\\  So $|z^{2}|+Re(az)+b=0$ has solutions if and only if $(f^{2}+g^{2})+(fd-eg)+b=0$ because $|z^{2}|=z\overline z$ Note that:  \begin{align}  |z^{2}|=z\overline z=f^{2}+g^{2}\\  \text{and}\\  Re(az)=fd-eg  \end{align}  So $|z^{2}|+Re(az)+b=0$ has solutions if and only if $(f^{2}+g^{2})+(fd-eg)+b=0$ because $|z^{2}|=z\overline z$  \end{proof}