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. 
 A: 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 $z_2 = \frac 12 - \frac i2 \sqrt 3$.
A: A comment on the uniqueness of the solution of achile hui: and @Martin R: and a generalization 
Take $f(z) = az^2 + bz + c$ with $a$, $b\ge 0$, $|f(z)|\le 1$ for $|z|\le 1$ ( enough for $|z|=1$) and such that $a+b=\frac{2}{\sqrt{3}}$, the maximum, 
With $\omega=e^{i \pi/3}$ we have 
$$a+b = \frac{f(\omega)-f(\omega^{-1})}{i \sqrt{3}}$$
so its absolute value does not exceed $\frac{2}{\sqrt{3}}$. Since we have equality we  conclude
$f(\omega)=i $  and  $f(\omega^{-1})=-i$. 
So we must have 
$$f(z) = \frac{2z^2 +4 z -1}{3\sqrt{3}} + \eta\cdot ( z^2-z+1) = f_0(z) + \eta\cdot ( z^2-z+1)$$ for some  $\eta$ real.
The function $|f_0(e^{i \theta})|$ achieves a maximum at $\theta = \pi/3$ so the tangent to the curve $f_0(e^{i\theta)}$ is horizontal. Since $f$ is also an extreme, the tangent to $f(e^{i\theta})$ at $\theta = \pi/3$ has the same direction. Recall that $\frac{d}{d\theta}  f(e^{i \theta})= i e^{i \theta} f'(e^{i \theta})$.
Therefore, we must have 
$$\frac{ \eta (2 \omega -1) }{4 \omega + 4} \in \mathbb{R}$$ and that implies $\eta = 0$. Therefore $f=f_0$. 
The plot of $\theta \mapsto \frac{2 e^{2 i \theta} + 4 e^{i \theta} -1}{3 \sqrt{3}}$ is a cardioid ( see pic )The fixed circle has center $(-\frac{1}{3\sqrt{3}},0)$ and radius $\frac{2}{3\sqrt{3}}$. The parametrization will match if we start rolling the circle from the right.
ADDED: After some calculations I found the  maximum of $s|a| + |b|$, where $s\ge 0$ is a constant. There are two cases: 


*

*$0\le s \le \sqrt{2}$, so $s = 2 \cos \theta$, where $\theta \in [\pi/4, \pi/2]$. Then  $\max (2 \cos\theta |a| + |b|) =\frac{1}{\sin \theta}$.

*If $s\ge \sqrt{2}$ then  $\max (s|a|+|b|)=s$.  
The inequality $2 \cos \theta |a| + |b| \le \frac{1}{\sin \theta}$ is proved in the same way by evaluating the polynomial 
$f$ at $\omega, \bar \omega = \cos \theta \pm i \sin \theta$.
If moreover $\theta \in [\pi/4, \pi/2]$, one checks that the value $\frac{1}{\sin \theta}$ is achieved for the polynomial 
$$f_{\theta}(z) = \frac{\cos \theta\cdot  z^2 - 2 \cos 2 \theta \cdot z + \cos \theta \cos  2 \theta }{2\sin^3 \theta}$$
$f_{\theta}$ was found so that $f(e^{\pm i \theta}) = \pm i$ and the curve $t \mapsto f(e^{\pm i t})$ has a horizontal tangent at $t = \phi$.
One checks that we have the equality
$$ 1- |f_{\theta}(e^{it})|^2 = \frac{- \cos (2 \theta) \cos^2 \theta\, (\cos t - \cos \theta)^2 }{\sin^6 \theta}$$
(so the  need for $\theta \ge \pi/4$ ). 
The uniqueness of the optimal polynomial (up to some phase changes) also holds. 
Note that for $\theta = \pi/4$ we get $\max \sqrt{2} |a| + |b| = \sqrt{2}$ achieved for the (essentially) unique polynomial $z^2$. 
If $s\ge \sqrt{2}$
we have 
$$s|a| + |b| = \frac{s}{\sqrt{2}} \cdot ( \sqrt{2}|a| + \frac{\sqrt{2}}{s}|b|)\le \frac{s}{\sqrt{2}} (\sqrt{2}|a|+|b|)\le \frac{s}{\sqrt{2}} \cdot \sqrt{2} = s$$
Therefore, the maximum of $s|a| + |b|$ is $s$ ( achieved for $z^2$). 
A: The minimization problem can be stated as
$$
\max_{z,a,b} \vert a\vert + \vert b\vert
$$
subject to
$$
\vert z \vert \le 1\\
\vert a z^2+b z + c \vert \le 1
$$
We will relax the problem to a more amenable form as
$$
\max_{z,a,b} \vert a\vert^2 + \vert b\vert^2
$$
subject to
$$
\vert z \vert = 1\\
\vert a z^2+b z + c \vert = 1
$$
according to this set of restrictions the maximum is located at the feasible region boundary
Forming the Lagrangian
$$
L = \vert a\vert^2 + \vert b\vert^2 + \lambda (\vert a z^2+b z + c \vert -1)
$$
After some cumbersome algebraic operations we get at
$$
\max_{z,a,b} \vert a\vert^2 + \vert b\vert^2 = \frac{1}{2}(1+\vert c\vert^2+2\vert c\vert)
$$
NOTE
$$
\vert z \vert = 1 \Rightarrow z = e^{i\phi}
$$
so the problem changes again to
$$
\max_{\phi,a,b} \vert a\vert^2 + \vert b\vert^2
$$
subject to
$$
\vert a e^{2i\phi}+b e^{i\phi} + c \vert = 1
$$
