Maximizing the value of $|z|$ where $az^2 + bz + c = 0$ Let $a,$ $b,$ $c,$ $z$ be complex numbers such that $|a| = |b| = |c| > 0$ and
$$az^2 + bz + c = 0.$$Find the largest possible value of $|z|.$

I immediately set up equations which told me that $$a_1^2 + a_2^2 = b_1^2 + b_2^2 = c_1^2 + c_2^2$$ and that $$(a_1 + a_2i)(z_1+z_2i)^2 + (b_1+b_2i)(z_1+z_2i) + (c_1 + c_2i).$$ However, I'm not sure how to move on from here. Can someone give me a push in the right direction?
 A: The roots of
$$z^2+e^{i\phi}z+e^{i\psi}=0$$
are
$$\frac{-e^{i\phi}\pm\sqrt{e^{i2\phi}-4{e^{i\psi}}}}2.$$
The largest value is when all terms add up maximally (by being aligned), giving

$$\frac{1+\sqrt{1+4}}2.$$

A: (It turns out this is a long version of Yves Daoust's answer.)
Let $r = |a| = |b| = |c| > 0$ and let $a = r \mathrm{e}^{\mathrm{i} \theta(a)}$, $b = r \mathrm{e}^{\mathrm{i} \theta(b)}$, and $c = r \mathrm{e}^{\mathrm{i} \theta(c)}$.  Notice that
$$  a z^2 + b z + c = 0  $$
if and only if
$$  \mathrm{e}^{-\mathrm{i} \theta(a)} a z^2 + \mathrm{e}^{-\mathrm{i} \theta(a)}b z + \mathrm{e}^{-\mathrm{i} \theta(a)}c = 0  \text{,}  $$
or, what is the same thing,
$$  r z^2 + r \mathrm{e}^{\mathrm{i} (\theta(b) - \theta(a))}z + \mathrm{e}^{\mathrm{i} (\theta(c) - \theta(a))} = 0  \text{.}  $$
In other words, we can rotate the plane containing the points $a$, $b$, and $c$ until $a$ lands on the positive real axis and the polynomial having the rotated coefficients has the same roots as the original polynoimal.
By the quadratic formula,
\begin{align*}
z &= \frac{-r \mathrm{e}^{\mathrm{i} (\theta(b) - \theta(a))} \pm \sqrt{r^2 \mathrm{e}^{2\mathrm{i} (\theta(b) - \theta(a))} - 4r\cdot r\mathrm{e}^{\mathrm{i} (\theta(c) - \theta(a))}}}{2r}  \\
    &= \frac{-r \mathrm{e}^{\mathrm{i} (\theta(b) - \theta(a))} \pm |r|\sqrt{\mathrm{e}^{2\mathrm{i} (\theta(b) - \theta(a))} - 4\mathrm{e}^{\mathrm{i} (\theta(c) - \theta(a))}}}{2r}  \\
    &= \frac{1}{2} \left( - \mathrm{e}^{\mathrm{i} (\theta(b) - \theta(a))} \pm \sqrt{\mathrm{e}^{2\mathrm{i} (\theta(b) - \theta(a))} - 4\mathrm{e}^{\mathrm{i} (\theta(c) - \theta(a))}} \right)  \\
\end{align*}
Let $u = \mathrm{e}^{\mathrm{i} (\theta(b) - \theta(a))}$ and $v = \mathrm{e}^{\mathrm{i} (\theta(c) - \theta(a))}$, so that $|u| = |v| = 1$ and
$$  z = \frac{1}{2} \left( -u \pm \sqrt{u^2 - 4v}   \right)  \text{.}  $$
Then
$$  |2z+u|^2 = |u^2 - 4v|  \text{.}  $$
Now the triangle inequality gives
$$  \min\{|u|^2 - 4|v|, 4|v| - |u|^2\} \leq |u^2 - 4v| \leq 4|v| + |u|^2  $$
and we can use our knowledge that $|u| = |v| = 1$ to obtain
$$  -3 \leq |u^2 - 4v| \leq 5  \text{.}  $$
We know moduli are nonnegative so $|u^2 - 4v| \in [0,5]$.  So \begin{align*}
|2z+u|^2 \in [0,5]  \\
|2z+u| \in [0,\sqrt{5}]  \text{.}
\end{align*}
Since $|u| = 1$, we have
$$  |2z| - 1 \leq |2z+u| \leq |2z|+1  \text{.}  
$$
From the left, $|2z| \in [1,1+\sqrt{5}]$, so $|z| \in [1/2,\frac{1+\sqrt{5}}{2}]$.  From the right, $|2z| \in [0,\sqrt{5} - 1]$, which does not give a larger upper bound.
Thus, an upper bound is $|z| \leq \frac{1+\sqrt{5}}{2}$.
Since we have used the triangle inequality separating dependent quantities ($-u$ and $u^2$ are not independent), we should verify that there is an assignment of the arguments of $u$ and $v$ that make the triangle inequality extremal.  Analyzing the first use, we require $v$ and $u^2$ be antiparallel.  Analyzing the second use, we require $-u$ and $u^2$ be antiparallel.  So, for $u$, we require an argument whose double points in the opposite direction, for instance $\pi$ is a possible argument for $u$.  Then $v$ is antiparallel to $u^2$ so is parallel to $u$.  Checking the polynomial
$$  z^2 - z - 1 = 0  \text{,}  $$
we find that it has the root $\frac{1 + \sqrt{5}}{2}$, so the upper bound we found is actually attained for particular values of $u$ and $v$, hence for particular values of $a$, $b$, and $c$, and therefore is the maximum modulus of $z$.
