Cosine of complex variable 
Show that $|\cos z|\leq2$  for all $z \in B(0,r)\subset C$ for some $r>0$ . Can we have $|\cos z|\leq 1$  for all $z \in B(0,r)\subset C$ for some $r>0$ ?

I have found the answer for the first part but can't think of a solution for the second question. For the first part
$$|\cos z|=\biggl|\frac{e^{iz}+e^{-iz}}{2}\biggr|$$
$$ =\biggl|\frac{e^{ix}*e^{-y}+e^{-ix}*e^{y}}{2}\biggr|$$
$$ \leq\biggl|\frac{e^{ix}*e^{-y}}{2}\biggr|+ \biggl|\frac{e^{-ix}*e^{y}}{2}\biggr| $$
$$ \text{But }\:|e^{-ix}|=|e^{ix}|=1$$
$$ =\biggl|\frac{e^{-y}}{2}\biggr|+ \biggl|\frac{e^{y}}{2}\biggr|$$
$$= \cosh y$$
$$\text{So }\: |\cos z|\leq \cosh  y$$
and since $\cosh 0=1$ we can find a neighborhood around the point $0$ such that $|\cos z|\leq 2$.
Since $\cosh x$ is a strictly increasing function I can't use it to prove that $|\cos z|\leq 1 $ for some neighborhood. Another method I can think of is taking partial derivative of $|\cos z|$ along $x$ and $y$-axes to check the direction in which it increases or decreases. But is there a simpler way for checking whether such a neighborhood exists ?

For $r>0$, consider a disc $B(0,r)$. Then at some point on imaginary axis $(x=0)$, you have
$f(y)=\cos (iy)=\cos hy=\frac{e^y+e^-y}{2}$
Since $f(y)$ attains its minimum value $1$ at $y=0$ and is an increasing function $( f'(y)>0)$ for $y>0$ hence $|f(y)|\ge 1$ (even for $y<0$). Thus $|\cos z|\le 1$ for all $z\in B(0,r)$ is impossible regardless of the choice of $r$.
