Complex analysis- poles vs. Zeros, etc.

[quasar_4]

I am having a hard time understanding the difference between poles and zeros, and simple poles versus removable poles. For instance, consider $$f(z)=\frac{z^2}{sin(z)}$$. we can expand sine into a power series and pull out a z, so doesn't that remove the singularity at z=0? Also, I don't see why n*pi would not also be removable since it doesn't seem to be a problem in the series expansion (but according to my graded homework, 0 is a zero and n*pi is a simple pole)... Can someone help me out here?

 

[Hurkyl]

we can expand sine into a power series and pull out a z, so doesn't that remove the singularity at z=0?

That's how you remove the singularity. But this operation produces a new (partial) function that is not f. (The difference being that this function is defined at 0 whereas f is not)

 

[quasar987]

But 0 is not a zero, it is a removable singularity! :/

 

[HallsofIvy]

But 0 is not a zero, it is a removable singularity! :/

?? What is your point? Hurkyl's point was that if f(z) has a "removable singularity" at $z_0$, yes, you can "remove" it but then you get a different function, g(z). g(z)= f(z) for all z except $z_0$. He never said anything about being a zero.

 

[quasar987]

My comment was in response to

(but according to my graded homework, 0 is a zero and n*pi is a simple pole)... Can someone help me out here?

HallofIvy.

 

[skook]

To try and sum up:
1) Cancel z top and bottom to show that the bottom term -> 1 as z -> 0. So that would remove the singularity and make the function analytic at zero.
1a) Because the bottom can't go to zero, the function must -> 0 when z -> 0. So there is a zero of the function at z = 0.
2) But if you don't cancel the z and stick with the original function, the sin(z) will vanish every time z -> n*pi and the function will go through the roof. So there are simple poles when z = n*pi.

hope this helps.