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Veva Garcia edited untitled.tex
almost 8 years ago
Commit id: 082e3e72370e221515650f446ba5ef6c40ed5547
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Then the right hand side $$ -b + \frac{1}{4}|a^2|\geq 0.$$
Thus, $\frac{1}{4}|a^2|\geq b$\\
Hence, $|a^2|\geq 4b$
Therefore, the equation $\left|z^2\right|+Re(az)+b=0$ has a solution if and only if $\left|a^2 \right| \geq 4b.$
\end{proof}
\end{problem}