# Infinite being removable singularity

1. Aug 5, 2012

### karlzr

1. The problem statement, all variables and given/known data
I am confused about the concept of removable singularity, when it comes to the infinite. Here are two examples in which infinite is claimed to be the removable singularity:
1, $f(z)=\frac{1+z^4}{z(1+z^2)^3}$;
2, $f(z)=sin\frac{1}{z-1}$
Actually, I don't even know why the infinite should be isolated singularity in the first place. Please explain in detail on the above examples, thanks!

2. Relevant equations

complex analysis, singularity

3. The attempt at a solution
I tried to write z in terms of 1/t, but failed to get the expected result.

Last edited: Aug 5, 2012
2. Aug 5, 2012

### Dick

Show what you tried. What was the 'expected result' you were looking for? Writing them as a function of 1/t you should be able to simplify to a function that's analytic at t=0. What's the problem?

3. Aug 5, 2012

### karlzr

To the first function $f(z)=\frac{1+z^4}{z(1+z^2)^3}$:
assume t=1/z, we can find:
$$f(1/t)=\frac{t^3(1+t^4)}{(1+t^2)^3}$$, which is regular at t=0. So I suppose f(z) should be regular at infinity, other than an isolated singularity.

4. Aug 5, 2012

### Dick

You've got it. f(z) is formally undefined at z=infinity but it has a perfectly well defined limit there. Just like f(1/t) is regular at t=0.

5. Aug 5, 2012

### karlzr

Does this mean that the infinity is always a singularity? If not, please give me an couterexample, thanks a lot!

6. Aug 5, 2012

### Dick

I would say so. You really can't substitute 'infinity' into functions. You can only take the limit as z->infinity. If you are dealing with some notion of 'extended' complex numbers like the Riemann sphere where you add a point at infinity, you might quibble about it. But I don't think that should worry you.

7. Aug 5, 2012

### karlzr

Thanks! My doubt on this issue has been cleared. Another problem:
consider the following three integrals:
$$I_1=\oint_{|z|=\pi}dz\frac{z+2}{(4z^2+1)(z-4)^2}$$
$$I_2=\oint_{|z|=\pi}dz\frac{z^2+2z}{(4z^2+1)(z-2)^2}$$
$$I_3=\oint_{|z|=\pi}dz\frac{z^3+2z^2}{(4z^2+1)(z-1)^2}$$
One of the these integrals vanishes. Explain which one and why, without calculating any residues.

8. Aug 5, 2012

### Dick

Now that you are comfortable with the idea of a function being regular at infinity, that should be a big clue. What happens if the function f(z) has all of its poles inside of the circle |z|=1 and is regular at infinity and you integrate around the circle? Think about the transformation z->1/z.

Last edited: Aug 6, 2012
9. Aug 6, 2012

### karlzr

I refer to my math book and there is a statement: "even if the infinity is not a singularity of f(z) at all, we cannot conclude $res(∞)=0$, as long as the $C_{-1}$in the expansion doesn't vanish".
At first, I think the third integral should vanish, since the residue at infinity vanishes and all three singularities lie within the contour. After some calculation, I am confused.
$$I_3=\oint_{|z|=\pi}\frac{z^3+3z^2}{(2z+i)(2z-i)(z-1)^2}$$.
I do the calculation with Mathematica and get the residues:
$res(i/2)=-19/100+2i/25; res(-i/2))=-19/100-2i/25; res(1)=11/25$
the sum does not vanish.
I expand the integrand and the coefficient $C_{-1}$ doesn't vanish.
As a matter of fact, the sum of the four residues is not zero.

10. Aug 6, 2012

### Dick

Look at what's happening for large z in that function. You've got a z^3 in the numerator and z^4 in the denominator. So for a large circle |z|=R the integrand is of the order 1/R and you are integrating over a circle of radius R. It's not going to zero fast enough to force the integral to vanish as R->inf. It does have a 'residue at infinity'. Your book is correct. I was oversimplifying. How about the second one?

Last edited: Aug 6, 2012