# Homework Help: Uniform convergence

1. Oct 6, 2012

### stephenkeiths

1. The problem statement, all variables and given/known data
I need to show that $f_{n}$=sin($\frac{z}{n}$) converges uniformly to 0.

2. Relevant equations
So I need to find K($\epsilon$) such that $\forall$$n \geq K$
|sin($\frac{z}{n}$)|<$\epsilon$

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

3. 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!

Last edited: Oct 6, 2012
2. Oct 6, 2012

### jbunniii

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

3. Oct 6, 2012

### 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| ????

Last edited: Oct 7, 2012
4. Oct 7, 2012

5. Oct 7, 2012

### 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
$$|\sin(z)| \leq |\sin(x) \cosh(y)| + |\cos(x) \sinh(y)| \leq |\cosh(y)| + |\sinh(y)|$$
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 $n \rightarrow \infty$.