Is it true that $\lvert \sin z \rvert \leq 1$ for all $z\in \mathbb{C}$? Is it true that $\left\lvert \sin z \right \rvert \leq 1$ for all $z \in \mathbb{C}$ ?
I think that is not true, can anyone help me?
 A: By Liouville's theorem the only bounded and entire functions are constant functions. An interesting deduction from Liouville's theorem is that non-constant entire functions must be unbounded. Hence $\sin z$, $\cos z$ being a non-constant entire function must be unbounded. 
A: Say we want $\sin z = 2$.
$$
2=\sin z = \frac{e^{iz}-e^{-iz}}{2i}
$$
if
$$
4i = e^{iz}+e^{-iz}.
$$
Multiply both sides by $e^{iz}$, getting
$$
4ie^{iz} = e^{2iz}+1.
$$
I.e.
$$
4iu = u^2 + 1.
$$
$$
u^2 - 4iu + 1 = 0.
$$
This is a quadratic equation
$$
au^2+bu+c=0
$$
with $a=1,\quad b= -4i,\quad c=1$.  The discriminant is
$$
b^2-4ac = -16-4 = -20 = -2^2\cdot5.
$$
So
$$
u = \frac{4i \pm2i\sqrt{5}}{2} = 2i \pm i\sqrt{5}.
$$
If we want
$$
e^{i(x+iy)}=e^{iz} = i(2+\sqrt{5})
$$
where $x$ and $y$ are real, then we need
$$
e^{-y} = 2+\sqrt{5}, \text{ so } y = -\log_e(2+\sqrt{5}), 
$$
and
$$
x = \frac \pi 2 \pm 2\pi n.
$$
$$
z= \frac \pi 2 -i\log_e(2+\sqrt{5}) + 2\pi n.
$$
So we've got $\sin z = 2$.
A: $$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}- \dots $$
$$\sin i=i\left( 1+\frac{1}{3!}+\frac{1}{5!}+ \cdots \right)$$
$$|\sin { i}|= 1+\frac{1}{3!}+\frac{1}{5!}+ \cdots >1$$
A: Liouville's theorem says "Every bounded entire function is a constant".
A: $\sin(z) = {1 \over 2i}(e^{iz} - e^{-iz})$, so for $z = -iy$ with $y$ real you have
$$|\sin(iy)| = \bigg|{e^{y} - e^{-y}  \over 2}\bigg|$$
So for large values of $y$ you have that $|\sin(iy)|$ is much larger than 1. 
