# 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?

• If it were true then the sine function wouldn't be very interesting nor useful ;) May 7, 2012 at 2:40
• Hint: $\sin\,ix=i\sinh\,x$, where $\sinh\,x=\dfrac{\exp\,x-\exp(-x)}{2}$ and consider the case where $x$ is a large real number. May 7, 2012 at 2:41

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

• This is like cracking nut with a sledgehammer Jul 5, 2012 at 10:50
• @Norbert, around here, we use the idiom "nuking mosquitoes"... :) Jul 5, 2012 at 11:21

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

• The second equation should read $4i=e^{iz}-e^{-iz}$. The error carries on from there. Oct 16, 2013 at 16:41
• Is it acceptable to edit others' answers to correct mistakes? Feb 23, 2015 at 18:36
• @EthanAlvaree : Probably in matters like this I'd say it is. However, if a student asks why $(d/dx)\sin x = tan x$, I would leave it intact, since the mistake is an essential part of the question. I'd put the correction in the answer. ${}\qquad{}$ Feb 24, 2015 at 1:04

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

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

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