# Limits equality, complex.

1. Dec 15, 2013

### binbagsss

I am trying to show {fn} converges uniformly on {z:|z|≤p}, where p is postiive real number.
and fn(z)=sin(z/n).

I am able to follow parts of my books method, but don't understand a couple of the inequalities...(my trouble lies in the inequalities rather than the main concepts involved in the method).

Sol:

I can see the pointwise limit is 0.

I am using the theorem that 'If for each n, we can find a sequence Sn, s.t |fn(z)-f(z)|≤Sn, and lim n→∞ Sn=0, then {fn} converges uniformly to f(z)'

So I am to consider: |Sin(z/n)-0|and looking for a Sn with the above properties.

so let z=x+iy and we attain |Sin(z/n)-0| = |Sin(x/n)cosh(y/n)+icos(x/n)sinh(y/n)| which is fine.

we than get |sin(x/n)cosh(y/n)+icos(x/n)sinh(y/n)|≤|sin(x/n)cosh(y/n)|+|sinh(y/n)|*
, which I am not 100% sure on.
I can see we have applied the triangle inequality. But on the right hand-term we have clearly applied another inequaliity , is this that |icos(x/n)| ≤1??

It then says for |z|≤p**:

|sin(x/n)cosh(y/n)|+|sinh(y/n)|≤(p/n)cosh(p/n)+sinh (p/n).

1) I think from * we have used that sin (x/n) ≈ (x/n) for small (x/n), but why don't we use a approximation sign rather than inequality?
2) MAIN question really - I'm not too sure how we have used ** ? haven't we set x=y=p. I'm having great difficulty understanding why we do this, I can see we have lost the absolute value signs...

Many Thanks for any help, really appreciated ! :)

Last edited: Dec 15, 2013
2. Dec 15, 2013

### Dick

They are using such facts as, i) if r is real, then |cos(r)|<=1, ii) if r is real then |sin(r)|<=|r|. The usual sorts of things you know about sin and cos as real functions. They are also using that if |x+iy|<=p then |x|<=p and |y|<=p.