On $[-1,1]$ we have Markov inequality for polynomials
$||P'||_{[-1,1]} \le (deg\ P)^2||P||_{[-1,1]} $
If $p$ is polynomial considered on $[a,b]$, then $q(x)=p(\frac{(b-a)x}{2}+\frac{a+b}{2})$ is a polynomial considered on [-1,1]. Using that and inequality on $[-1,1]$ we get
$||p'||_{[-1,1]} \le \frac{2}{|b-a|}(deg\ p)^2||p||_{[-1,1]} $
But what if $a$ and $b$ are complex numbers? Is it the same?
