who know how to prove the estimate $Re(\frac{1}{\zeta^2})\gt\frac{4}{81t^2}$ (pr something like this)?
You have to prove that if $\zeta=t+(\frac{t}{2})\exp(i\theta)$ then $Re(\frac{1}{\zeta^2})\geq\frac{4}{81t^2}$.
Thanks in advance.
who know how to prove the estimate $Re(\frac{1}{\zeta^2})\gt\frac{4}{81t^2}$ (pr something like this)?
You have to prove that if $\zeta=t+(\frac{t}{2})\exp(i\theta)$ then $Re(\frac{1}{\zeta^2})\geq\frac{4}{81t^2}$.
Thanks in advance.
Presumably $t$ and $\theta$ are real. By homogeneity we may as well take $t=1$, so $\zeta = 1 + \dfrac{\cos(\theta)+i\sin(\theta)}{2}$.
I get $\text{Re} \frac{1}{\zeta^2} = 4\,{\frac {3+4\,\cos \left( \theta \right) +2\, \cos^2 \left( \theta \right) }{25+40\,\cos \left( \theta \right) +16\, \cos^2 \left( \theta \right) }} $ Now $3 + 4 \cos(\theta) + 2 \cos^2 (\theta) = 1 + 2 (1 + \cos(\theta))^2 \ge 1$ while $25 + 40 \cos(\theta) + 16 \cos^2(\theta) \le 25 + 40 + 16 = 81$.
EDIT: How to get that $\text{Re} \dfrac{1}{\zeta^2}$? Note that $\text{Re}\frac{1}{z} = \text{Re} \frac{\overline{z}}{|z|^2} = \frac{\text{Re} \;z}{|z|^2}$ Do this for $z = \zeta^2$, expanding and simplifying as necessary.