if $|az^2+bz+c|\le 1$, find the maximum of $|a|+|b|$ Let $a,b,c$ be complex numbers, and $f(z)=az^2+bz+c$, such that if the complex number $|z|\le 1$, then we have $|f(z)|\le 1$. Find the maximum of $|a|+|b|$.
if we $a,b,c$ be real and $z$ be real, I can find the maximum is $2$,because $f(z)=2z^2-1$ such  it. and
$$a=\dfrac{1}{2}[f(1)+f(-1)]-f(0),b=\dfrac{1}{2}[f(1)-f(-1)]$$
so
$$|a|+|b|=\dfrac{1}{2}|f(1)+f(-1)-2f(0)|+\dfrac{1}{2}|f(1)-f(-1)|\le 2$$But for complex, maybe this answer is also $2?$,I'm not sure. 

By replacing $f(z)$ with
$$
 \tilde f(z) = e^{i\phi} f(e^{i\psi} z) =  e^{i(\phi+2\psi)}a z^2 + e^{i(\phi+\psi)}b z + c = \tilde a z^2 + \tilde b z + c
$$
with suitably chosen $\phi, \psi \in \Bbb R$ we can assume that both
$a$ and $b$ are real and $\ge 0$.
Now let $\omega = e^{i\pi/3} = \frac 12 + \frac i2 \sqrt 3$ be the primitive $6^{\text{th}}$ root of unity. Then $\omega^2 = \omega - 1$ and $ \omega^{10} = \omega^5 - 1$.
Therefore
$$
 f(\omega) = (a+b)\omega + c - a \\
 f(\omega^5) = (a+b)\omega^5 + c - a
$$
which implies
$$
  a + b = \frac{f(\omega)- f(\omega^5)}{\omega - \omega^5}
 = \frac{f(\omega)- f(\omega^5)}{i \sqrt 3} \, .
$$
Now use that $|f(z)| \le 1$ on the unit circle, this gives
the estimate
$$
 |a| + |b| = a + b \le \frac{2}{\sqrt 3} \, .
$$
And this is the actual maximum. Credit for the following example
goes to achille hui: Let  $p(z) = 2z^2+4z - 1$, then
$$
\begin{align}|p(e^{it})|^2 &= |2e^{it} - e^{-it} + 4|^2 = |\cos(t) + 4 + 3i\sin(t)|^2\\  &=(\cos(t)+4)^2 + 9\sin(t)^2 = 25 +8\cos(t)(1-\cos(t))\\ &\le 25 + \frac{8}{4} = 27 \, ,\end{align} 
$$
so that
$$
 f(z) = \frac{2z^2 + 4z - 1}{3\sqrt{3}}
$$
satisfies $|f(z)| \le 1$ on the unit circle and therefore – due to
the maximum principle – for all $z$ in the unit disk.

Remark: This is how I came up with above approach: In order to
compute $a+b$ from two equations 
$$
 f(z_1) = a(z_1^2-z_1) + (a+b)z_1 + c \\
 f(z_2) = a(z_2^2-z_2) + (a+b)z_2 + c 
$$
we need different $z_1, z_2$ with $z_1^2-z_1 = z_2^2-z_2$,
or $z_1 + z_2 - 1 = 0$. Also $z_1, z_2$ should be of absolute
value $\le 1$, and their difference as large as possible.
This eventually led to the choice $z_1 = \frac 12 + \frac i2 \sqrt 3$ and $z_2 = \frac 12 - \frac i2 \sqrt 3$.
