Infinite being removable singularity

[karlzr]

Homework Statement

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!

Homework Equations

complex analysis, singularity

The Attempt at a Solution

I tried to write z in terms of 1/t, but failed to get the expected result.

 

[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?

 

[karlzr]

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?

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.

 

[Dick]

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.

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.

 

[karlzr]

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.

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

 

[Dick]

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

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.

 

[karlzr]

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.

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.

 

[Dick]

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.

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.

 

[karlzr]

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.

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.

 

[Dick]

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.

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?