Find the number of roots for the following equation for $|z| < 2$, $z\in \Bbb C$: $$ z^2 - \cos z=0 $$
The reasoning below is based on using Rouche's theorem. So basically I picked two functions $f(z)$ and $\phi(z)$ such that: $$ f(z) = z^2\\ \phi(z) = \cos z $$
Now I need to show that $|f(z)| > |\phi(z)|$ for $|z| = 2$. My main issue here is proving that statement.
Consider $|z^2|$, clearly the absolute value is $4$. For $|\cos z|$ and $y\in[-2,2]$: $$ \begin{align} |\cos z| &= \left|\frac{e^{iz} + e^{-iz}}{2}\right| \\ &= {1\over 2}\left|e^{ix - y} + e^{-ix+y}\right| \\ &\le {1\over 2}\left(\left|e^{ix}\right|\left|e^{-y}\right| + \left|e^{-ix}\right|\left|e^{y}\right|\right)\\ &={1\over 2}\left(e^{-y}+e^y\right) \end{align} $$
Now $|\cos z|$ is symmetric with respect to $x = 0$ so we might consider only one case: $y \in [0;2]$. Take a look at the following equation: $$ \begin{align} {1\over 2}\left(e^{-y} + e^y\right) &= 4 \\ e^{-y} + e^y &= 8 \ \ | \times e^y \\ e^{2y} - 8e^y + 1 &= 0 \end{align} $$ Solve for $e^y$: $$ e^y = 4\pm \sqrt{15} $$
Approximate calculations show that: $$ \begin{align} y = \ln(4+\sqrt{15}) &\approx 2,06343... > 2\\ y = \ln(4-\sqrt{15}) &\approx -2,06343... < -2 \end{align} $$
Finally since ${1\over 2}\left(e^{-y}+e^y\right)$ is incresing for $y \in [0,2]$ we have that: $$ |\cos z| = {1\over 2}\left(e^{-y}+e^y\right) < 4, \forall y\in[0,2] $$
This means the equation has two roots (with multiplicities) in $|z| < 2$.
The question here is how do I show $$ \ln(4+\sqrt{15}) > 2\\ \ln(4-\sqrt{15}) < -2 $$ Also, I would appreciate it if someone could show a simpler solution.
You already figured out that for $z=x+iy$ $$ |\cos(z)| \le \cosh(y) $$ and that $\cosh(y)$ is increasing in $y$ for $y \ge 0$. Therefore $$ |\cos(z)| \le \cosh(2) = \frac{e^2+e^{-2}}{2} $$ for $|z| =2$, and it suffices to show that $e^2+e^{-2} < 8$: $$ e^2+e^{-2} < 2.8^2 + \frac 1{2.5^2} = 7.84 + 0.16 = 8 \, . $$
It follows that $z^2 - \cos(z)$ and $z^2$ have the same number of zeros in $|z| < 2$ (counting multiplicities), that is $2$.