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.

Let \(a \in \mathbb{C}\), \(b \in \mathbb{R}\) and \(z \in \mathbb{C}\). Then \(z=x+yi\) and \(a=c+di\) for some \(x,y,c,d \in \mathbb{R}\).
Then we have, \[\left|z^2\right|+Re(az)+b=0,\]
By substitution we get the following: \[x^2+y^2 + Re((c+di)(x+yi))+b =0\]
By algebra, \[x^2+y^2 + Re(cx+cyi+dxi+dyi^2)=-b\]
\[x^2+y^2 + Re(cx-dy+i(dx+cy))=-b\]
\[x^2 +y^2 + cx-dy = -b\]
\[(x^2+cx) +(y^2-dy)=-b\]
Then by completing the square, \[(x+\frac{1}{2}c)^2 + (y-\frac{1}{2}d)^2 = -b + \frac{1}{4}(c^2+d^2)\]
Thus, \[(x+\frac{1}{2}c)^2 + (y-\frac{1}{2}d)^2 = -b + \frac{1}{4}|a^2|\] 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, \(|a^2|\geq 4b.\)
Assume \(|a^2|\geq 4b.\) Then \(\frac{1}{4}|a^2|\geq b.\) Thus, \(\frac{1}{4}(c^2+d^2)\geq b.\)

Therefore, the equation \(\left|z^2\right|+Re(az)+b=0\) has a solution if and only if \(\left|a^2 \right| \geq 4b.\)

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