Show that $|z^2| + Re(a \cdot z) +b = 0$ has only solutions for $|a^2|\geq 4b$

Question

The Question: show that $|z^2| + Re(a \cdot z) +b = 0$ only has a solution for $|a^2|\geq 4b$ $a \in \Bbb{C}$ $b \in \Bbb{R}$ Question 1.10

If I substitute $z=r_z e^{i\phi_z}$ and $a=r_a e^{i \phi_a}$

I get $r_z^2 + Re(r_z r_a e^{i(\phi_a+\phi_z)}) + b= 0$

$r_z^2 + r_z r_a cos(\phi_a+\phi_z) + b= 0$

Since $r_z$ has to be real, the root of $(r_a cos(\phi_a+\phi_z))^2 - 4b$ has to be real. Therefore $(r_a cos(\phi_a+\phi_z))^2 - 4b \geq 0$

But I have no idea how to get from this to $|a^2|\geq 4b$.

Did I go down the wrong path, did I make a mistake or am I just missing a last step?

Thanks

Answer 1

Hint The key observation is that each term can be written compactly in terms of an expression and its conjugate. Expanding gives $$(z + c)\overline{(z + c)} = z \bar{z} + z\bar{c} + \bar{z} c + c \bar{c} = |z|^2 + 2 \Re(z \bar{c}) + |c|^2 . $$

We see that we can match the second term with the second term on the l.h.s. of the given equation by setting $c := \frac{1}{2} \bar a$. Then, we can write the given equation as $$\left(z + \tfrac{1}{2} \bar a\right)\overline{\left(z + \tfrac{1}{2} \bar a\right)} - \left(\tfrac{1}{4} |a|^2 - b\right) = 0.$$