Problem 1.10

Fix an $$a\in\mathbb{C}$$ and $$b\in\mathbb{R}$$. Show that the equation
$$|z^{2}|+Re(az)+b=0$$ has a solution if and only if $$|a^2|\geq4b$$.

Let $$a\in\mathbb{C}$$,$$b\in\mathbb{R}$$, and $$z\in\mathbb{C}$$.
Then $$a=d+ei$$ and $$z=f+gi$$ for some $$d,e,f,g\in\mathbb{R}$$.
Note that: $|z^{2}|=z\overline z=f^{2}+g^{2}$
and $Re(az)=fd-eg$

So $$|z^{2}|+Re(az)+b=0$$ has solutions \begin{aligned} &\iff (f^{2}+g^{2})+(fd-eg)+b=0\\ &\iff f^{2}+fd+g^{2}-eg+b=0\\ &\iff (f+\frac{1}{2}d)^{2}-\frac{1}{4}d^{2}+(g-\frac{1}{2})^{2}-\frac{1}{4}e^{2}+b=0 \text{ (By completing the square)}\\ &\iff (f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}d^{2}+\frac{1}{4}e^{2}-b\\ &\iff(f+\frac{1}{2}d)^{2}+(g-\frac{1}{2}e)^2=\frac{1}{4}(d^{2}+e^{2})-b\\\end{aligned} The above equation is a circle with radius $$r=\sqrt{\frac{1}{4}(d^{2}+e^{2})-b}$$, so $$r^{2}$$ must be greater than or equal to zero in order for a solution to exist. Thus, \begin{aligned} \frac{1}{4}(d^{2}+e^{2})-b\geq0 &\iff(d^{2}+e^{2})\geq4b\end{aligned} And since $$a=d+ei$$, $$|a^{2}|=a\overline a=d^{2}+e^{2}$$. Thus a solution to $$|z^{2}|+Re(az)+b=0$$ exists if and only if $$|a^2|\geq4b$$.