deletions | additions       diff --git a/untitled.tex b/untitled.tex  index 2547ead..b3c1a30 100644  --- a/untitled.tex  +++ b/untitled.tex 
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Notice the left hand side $$(x+\frac{1}{2}c)^2 + (y-\frac{1}{2}d)^2 \geq 0$$  Then the right hand side $$ -b + \frac{1}{4}|a^2|\geq 0.$$  Thus, $\frac{1}{4}|a^2|\geq b$\\  Hence, $|a^2|\geq 4b$ 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}