Math Is Fun Forum

coffeeking

Member

Registered: 2007-11-18

Posts: 44

More Questions...

Q1:
Evalute

Where C(0,6) is the positive oriented circle |z|=6 and

I understand that for this question we need to use the Argument Principle in which

However, the solution are given something like:

Zeros in

are:

(order 3)

(order 1)

Therefore

My question is, why as for the sin(z) term we only take 0, pi and -pi? If we are considering only the principal value of argument i.e (-pi,pi] we should only take 0 and pi and not -pi since it does not lies in the interval. Anyone has any idea?

Q2:

Does there exist an entire function f such that

for all positive integers n? Justify your answer.

Alright as for this, the solution is given as:

Consider

then

for all positive integers n.

Now

iff

Therefore if z is a zero of

,

and

Hence g(z) has no poles in D(0,1) and so g is analytic on D(0,1).

Now

for all positive integers n and

in D(0,1) with

.

It follows that for any analytic function f with the given property,

.

However, limit of g at z=-1 does not exist implies that the limit of f at -1 does not exists and so f cannot be analytic at -1 and so f cannot be an entire function. Hence such entire function does not exist.

Well, the thing that I don't understand about this question is that since the function f is defined on positive integers n the domain and image set of f should all be positive so how could we possibly find limit of f at -1 at the very first place?

Furthermore, I don't understand why

implies that

.

Thanks a dozen in advance.

Last edited by coffeeking (2009-05-19 02:51:52)

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: More Questions...

1. The argument has nothing to do with this. The sine function has three zeros inside the circle C(0,6).

These are at 0 and

. If the circle had radius 7 instead we would need to consider the zeros at

as well.

2. I think that f is a hypothetical entire function such that

for all positive integers n.
Since it is entire it must be defined on the whole complex plane.

implies that

which clearly implies that

.
The point is that the poles of g do not lie in the open unit disk.

coffeeking

Member

Registered: 2007-11-18

Posts: 44

Re: More Questions...

Hi Avon, thanks for the help I appreciate it . Anyway, as for the last part of my question I do understand that

implies that

, which is enough to show that the poles of g do not lie in the open unit disk. However, to my understanding

will never implies

because

is exactly equal to 1 and never larger.

Please advise.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: More Questions...

The inequality is weak, so the implication is fine.
It's true because 1 ≥ 1.

---

Why did the vector cross the road?
It wanted to be normal.

coffeeking

Member

Registered: 2007-11-18

Posts: 44

Re: More Questions...

Thanks. So as to clear thing up if we are given

are we also right to say that

? and also,

implies

?