Maximum length of a complex number

Divide through by $a$ to get $z^2 + bz + c = 0, \, |b| = |c| = 1.$ So then $\begin{aligned} |z|^2 = |bz+c| &\le |b||z| + |c| = |z|+1 \\ |z|^2-|z|-1 &\le 0. \end{aligned}$ The roots of this equation are $\phi$ and $1/\phi,$ where $\phi = \frac{1+\sqrt{5}}2,$ so we get immediately that $\frac1{\phi} \le |z| \le \phi.$ Clearly equality holds on the left with $b=1, c=-1,$ and equality holds on the right with $b=c=-1.$ This gives us the maximum and minimum values of $|z|$ immediately.

WLOG let $a=1$ . then $z^2+bz+c=0$ for phase $b,c$ . we have $\large z= \dfrac{-b\pm\sqrt{b^2-4c}}{2} = \dfrac{b}{2} \left(-1\pm \sqrt{1-4\dfrac{c}{b^2}}\right) \to |z| =\left|\dfrac{b}{2}\right|\left|-1\pm \sqrt{1-4\dfrac{c}{b^2}}\right|= \dfrac{1}{2}\left|-1\pm \sqrt{1-4x}\right|$ with $x=\dfrac{c}{b^2}$ also being a phase. so we have: $\large z=\dfrac{1}{2}\left|-1\pm \sqrt{1-4x}\right|\leq \dfrac{1}{2} \left(|-1| +\left|\pm \sqrt{1-4x}\right| \right)=\dfrac{1+\sqrt{\left|1-4x\right|}}{2}\leq \dfrac{1+\sqrt{\left|1\right|+\left|-4x\right|}}{2}= \boxed{\dfrac{1+\sqrt{5}}{2}}$ with the equality case being $x=-1\to b^2=-c$ .

notice that if $z$ satisfies the properties given in the problem, so does $\dfrac{1}{z}$ , from this we can find the minima.