Fix $a\in\mathbb{C}$ and $b\in\mathbb{R}.$ Show that the equation $|z^2|+\text{Re}(az)+b = 0$ has a solution iff $|a^2|\geq 4b.$ When solutions exist, show the solution set is a circle.
A seemingly easy problem. Letting $z=x+yi$ and $a = c+di$ the equation can be rewritten as $$x^2+y^2+cx-dy+b = 0.$$
I don't know how to proceed.
On
You are almost done. Complete the square on $x$ and $y$ to get $$ (x+c/2)^2+(y-d/2)^2=c^2/4+d^2/4-b$$
This is an equation of a circle iff RHS is positive.
You can take over from here .
On
Alt. hint: using that $\,|w|^2 = w \bar w\,$ and $\,2\operatorname{Re}(w)=w+\bar w\,$ :
$$ 0 = z\bar z + \frac{a z + \bar a \bar z}{2} \color{red}{+\frac{a \bar a}{4} - \frac{a \bar a}{4}} + b = \left(z + \frac{\bar a}{2}\right)\left(\bar z + \frac{a}{2}\right) - \frac{|a|^2}{4} + b \\[10px] \iff\quad \left|z+\frac{\bar a}{2} \right|^2 = \frac{|a|^2-4b}{4} $$
It is a matter of completing squares: \begin{align*} & x^{2} + y^{2} + cx - dy + b = 0\Longleftrightarrow \left(x^{2} + cx + \frac{c^{2}}{4}\right) + \left(y^{2} - dy + \frac{d^{2}}{4}\right) + b - \frac{c^{2} + d^{2}}{4} = 0 \\ &\therefore \left(x+\frac{c}{2}\right)^{2} + \left(y - \frac{d}{2}\right)^{2} = \frac{c^{2}+d^{2}}{4} - b \geq 0 \Longleftrightarrow c^{2} + d^{2} \geq 4b \Longleftrightarrow |a|^{2} \geq 4b \end{align*}