Prove or disprove: $|\sin z|<1$ for all $z \in \mathbb C$

Prove or disprove: $|\sin z|<1$ for all $z \in \mathbb C$

I took some good advice and try $z=iy$ for $y\in \mathbb R$, so

$$\sin z= \sin iy = \frac{e^{i(iy)}-e^{-i(iy)}}{2i}$$ $$=\frac{e^{-y}-e^y}{2i}$$ $$=\frac{i(e^y-e^{-y})}{2}$$

Now say $y=5$

$$\sin z=\frac{i(e^5-e^{-5})}{2}=74.203i$$

so $|\sin z|=74.203 >1$. And that's it?

• Hint: what if $z=iy$ where $y$ is real? Feb 12, 2015 at 23:22
• Your inequality is on complex numbers, and in $\mathbb{C}$ you do not have $-1\leq \sin z\leq 1$, and indeed you do not have strict inequality. You can read here a bit about complex functions: milefoot.com/math/complex/functionsofi.htm Feb 12, 2015 at 23:24
• You are right: the simplest way to disprove this is to take $z = \frac{\pi}{2}$. However, I think you or your professor has made a typo, and actually wants to look at $|\sin(z)|\leq 1$. For this, the other answers suffice. Feb 12, 2015 at 23:33

7 Answers

The first thing you really ought to think about is your statement $$-1\le\sin z\le1$$ for complex $z$. You should realise that since the sine of a complex number is in general complex, this statement is not so much wrong as meaningless! There is no useful way of defining one complex number to be less than or greater than another.

As for the specific question you asked, if you simplify $$|\sin(iy)|$$ where $y$ is real, you should be able to show that it can be as large as you like and does not have to be less than $1$.

Good luck!

To kill a mouse with a cannon --- the only bounded holomorphic functions are constants.

The inequality you wrote has only been proven when $z$ is real.

The definition of $\sin z$ which extends it to complex numbers is $$\sin z=\frac{e^{iz}-e^{-iz}}{2i}\qquad z\in\mathbb C$$ So, are there any values of $z$ that can make the above expression very large? For example, for what values of $z$ is $e^{iz}$ large?

The statement in indeed false. But also $|\sin z| \le 1$ would be false; in fact $\sin z$ is not bounded if $z \in \mathbb C$.

This can be shown in several ways, for example by writing the $$\sin z = \frac{e^{iz} - e^{iz}}{2i}$$ and choosing an imaginary $z$.

I don't know if you know about holomorphic function, but note that the Liouville theorem implies that every bounded entire function must be constant. Since $\sin z$ is entire, if it was bounded then it would have to be constant but that is clearly false.

This is admittedly using a nuke to kill a fly though :-)

Well considering z=pi/2 would do the job were here z is a complex number where its imaginary part is zero! Dont forget that R is a subset of C

Just see this.

$$\sin(z) = \sin(x+iy) = \sin(x)\cosh(y)+i\cos(x)\sinh(y).$$

Can you bound it?

Well, we obviously know that $-1 \leq \sin (x) \leq 1$ for all $x \in \mathbb{R}$. But in $\mathbb{C}$ we have the introduction of the imaginary axis. I would suggest you try a few $z$'s along the imaginary axis. You might get something more substancial.