The problem is: for complex number $z=re^{i\theta}$, derive an inequality between $r$ and $\theta$ such that $$|\cos(2z)|<1$$
In my solution, I found that if $z=a+bi$, then the following is satisfied: $$\cos(4a)<2-\cosh(4b)$$
However, when I substitute in $a=r\cos(\theta)$ and b=$r\sin(\theta)$, the inequality becomes extremely complicated.
Since the problem is so simple, I believe that there is a solution that is in the form $f(r)<g(\theta)$. But I failed to obtain it.
Could anyone give a simpler solution?
Any help will be appreciated.
Expanding the complex $\cos$: $$ 2\cos(re^{i\theta}) = (e^{2r\sin\theta} + e^{-2r\sin\theta})\cos(2r\cos(\theta)) + i\sin(2r\cos(\theta)), $$ $$|\cos(re^{i\theta})| = |\cosh({2r\sin\theta})|, $$ And $\cosh$ is positive in $\Bbb R$ and $< 1$ in...