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?

Liouville's theorem says "Every bounded entire function is a constant".

$\sin(z) = {1 \over 2i}(e^{iz} - e^{-iz})$, 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.

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.

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 $\sin z = 2$.

$$\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}- \dots $$

$$\sin i=i\left( 1+\frac{1}{3!}+\frac{1}{5!}+ \cdots \right)$$

$$|\sin { i}|= 1+\frac{1}{3!}+\frac{1}{5!}+ \cdots >1$$