Prove Uniform Convergence of f_n=sin(z/n) to 0

[stephenkeiths]

Homework Statement

I need to show that f_{n}=sin(\frac{z}{n}) converges uniformly to 0.

Homework Equations

So I need to find K(\epsilon) such that \foralln \geq K
|sin(\frac{z}{n})|<\epsilon

I'm trying to prove this in an annulus: \alpha\leq |z| \leq\beta

The Attempt at a Solution

I'm having trouble because no matter what I choose for K I can't get the epsilon to come out.
I'm trying something like K(\epsilon)=\frac{1}{\alpha\epsilon}.

My problem is that I can't say that sin(\frac{z}{n})<sin(\frac{\alpha}{n})

Which is how I've been doing these uniform convergence ones (recasting in terms of \alpha instead of z).

Anyways I was hoping I could get some help on how to proceed.

Thanks!

 

[jbunniii]

Try using the fact that |\sin(x)| \leq |x| for all x.

 

[stephenkeiths]

Does that work for complex numbers too? It strikes me that if y=Im(z), then sin(z)<=cosh(y) ??

Writing z as z=x+iy, then sin(z)=\frac{e^{iz}-e^{-iz}}{2i} becomes: \frac{1}{2i}(e^{-y+ix}-e^{y-ix}) Then taking the magnitude and using the triangle inequality, it seems it'd come to \frac{1}{2}|e^{y}|+\frac{1}{2}|e^{-y}|

So then I can't use |sin(z)|\leq|z| ?

 

[stephenkeiths]

Any suggestions? Or comments?

 

[jbunniii]

Sorry, I didn't notice you were working with complex numbers. In that case if we write z = x + iy, then
\sin(z) = \sin(x) \cosh(y) + i \cos(x) \sinh(y)
so
|\sin(z)| \leq |\sin(x) \cosh(y)| + |\cos(x) \sinh(y)| \leq<br /> |\cosh(y)| + |\sinh(y)|
Therefore,
|\sin(z/n)| \leq |\cosh(y/n)| + |\sinh(y/n)|
So the problem reduces to showing that cosh(y/n) and sinh(y/n) converge uniformly to zero in the annulus as n \rightarrow \infty.