Let $ \lambda \in \mathbb{C} $ be any complex number, $ R > 0 $ some positive real. I want to find maxima of absolute value of some polynomial on the circle:
$$ \max (|x-1| |x-\lambda|) = \max| x^2 - (\lambda + 1)x + \lambda | = \max|P(x)|, $$ where $ |x| = R\ $ and $\ P(x) = x^2 - (\lambda + 1)x + \lambda $.
It is not hard to find maximal value for concrete values of $R$, but I want it as function of $R$ and $\lambda$. I tried two things.
Firstly, polar coordinates: $ x = R e^{i \phi} $, $\lambda = L e^{i\psi} $. After some calculations I got the following equation:
$$ (L^2 + R^2) \sin(\phi) + L(1 + R^2) \sin(\phi - \psi) - 2LR\sin(2\phi - \psi) = 0. $$
We have to solve it for $\phi$. I don't know, how to do it (neither does Wolframalpha).
Secondly, I notice that if $x_m$ is the point of maxima on the circle, then
$$ Arg \frac{P(x_m)}{P'(x_m)} = Arg (x_m). $$
The reason is that in other way we could step a little inside the cirle and increase the $ |P(x)| $ (it is almost obvious if you draw the picture). But that would be contrary to the maximum modulus principle ($ P(x) $ is holomorphic). That gives us folowing:
$$ \frac{x_m^2 - (\lambda+1)x_m + \lambda}{2x_m^2 - (\lambda+1)x_m} \in \mathbb{R}_+. $$ And again, I do not know, what can you do next.
Maybe there is some geometry approach?