# Simple inequality with complex number

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 $Re(\frac{1}{\zeta^2})\geq\frac{4}{81t^2}$.

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What is the counterexample? Or what is the relation between $\zeta$ and $t$? – Robert Israel Nov 15 '12 at 22:53
I think now is better! – TheStudent Nov 15 '12 at 23:03

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.
How do you go from $\zeta$ to $Re(\frac{1}{\zeta^2})$? Could you explain? – TheStudent Nov 16 '12 at 0:16
Right. I calculate $\zeta^2$ and $\vert\zeta^2\vert^2$, then use $Re\frac{1}{\zeta^2}=\frac{Re \zeta^2}{\vert \zeta^2 \vert^2}$ and the equality $\cos 2\theta = 2 \cos^2\theta -1$. It works! Thank you. – TheStudent Nov 16 '12 at 9:43