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 $$\left|a^2 \right| \geq 4b.$$