Let, $C(T)$ be the semi-circle centered at $c(>0)\in \Bbb R$ and radius $T$ lying to the left of the vertical segment. If $z=x+iy\in C(T)$ then for all large $T$ we have, $\displaystyle |z(z+1)|\ge \frac{T^2}{2}$.
My book says this. I think it is trivial one, but I'm unable to prove this. I've tried by putting $z=x+iy$ and also $z=c+Te^{i\theta}$. But I'm not getting the result. Any help?
$|z| \geq T-c$ so $|z(z+1)| \geq (T-c)(T-c-1) >(T-c-1)^{2}>\frac {T^{2}} 2$ for $T$ sufficiently large. In fact this is true for $T > \frac {1+c} {1-\frac 1 {\sqrt 2}}$