$f \colon \mathbb C \rightarrow \mathbb C$ be defined by $f(z)=\cos z.$ I was thinking about the problem that says:   

Let $f \colon \mathbb C \rightarrow \mathbb C$ be defined by $f(z)=\cos z.$ Then which of the following options is correct?
  $1.|f(z)| \leq 1$
  $2.|f(z)| \leq \pi$
  $3.|f(z)| \leq |z|$
  $4. f$ is unbounded.  

My Attempt: option $1$ is clearly false since $\cos i=f(i)=1+\frac{1}{2!}+\frac{1}{4!}+\cdots >1.$ But I am not sure about the other options.
Another observation: Also,I see $f(z)$ is analytic on $\mathbb C$ and it is non-constant. So ,It must be unbounded( by Liouville's theorem).So option $4.$ looks right.Am I right ?
Thanks in advance for your time.

$4.\cos z$ is unbounded.
We know that $\cos (iy)=\cosh y$ and $\sin(iy)=i \sinh y $
$$\cos z=\cos x \cosh y-i \sin x \sinh y$$
so, we get
$$|\cos z|^2=\cos^2 x+ \sinh^2 y$$
now it is easy to see that $\cos z$ is unbounded.Can u see it? 
