1. ## True or false about analycity

True or false? Justify.
Let $\displaystyle z_0$ be an isolated singularity of f(z) such that $\displaystyle Res(f,z_0)=0$. Then $\displaystyle z_0$ is a removable singularity and thus $\displaystyle f(z)$ is analytic in a neighborhood of $\displaystyle z_0$.
Attempt: True. I know that if $\displaystyle z_0$ is a removable singularity, then f can be extended to an analytic function by giving a value to $\displaystyle f(z_0)$. Thus I also believe that f is bounded if its singularity is removable.
Though formally I don't know how to prove it. If $\displaystyle f(z)=\sum _{n=-\infty}^{+\infty} a_n (z-z_0)^n$ (I know I assume that f has a Laurent series representation, I'm not sure I can assume that) then I'm told that $\displaystyle \lim _{z \to z_0} \sum _{n=-\infty}^{+\infty} a_n (z-z_0)^{n+1}=0$. I'm not going far with this.
Any idea?

2. Originally Posted by arbolis
True or false? Justify.
Let $\displaystyle z_0$ be an isolated singularity of f(z) such that $\displaystyle Res(f,z_0)=0$. Then $\displaystyle z_0$ is a removable singularity and thus $\displaystyle f(z)$ is analytic in a neighborhood of $\displaystyle z_0$.
Attempt: True. I know that if $\displaystyle z_0$ is a removable singularity, then f can be extended to an analytic function by giving a value to $\displaystyle f(z_0)$. Thus I also believe that f is bounded if its singularity is removable.
Though formally I don't know how to prove it. If $\displaystyle f(z)=\sum _{n=-\infty}^{+\infty} a_n (z-z_0)^n$ (I know I assume that f has a Laurent series representation, I'm not sure I can assume that) then I'm told that $\displaystyle \lim _{z \to z_0} \sum _{n=-\infty}^{+\infty} a_n (z-z_0)^{n+1}=0$. I'm not going far with this.
Any idea?
What about the function $\displaystyle \displaystyle f(z)=e^{1/z}-\frac1z = 1+\sum_{n=2}^\infty \frac{1}{n!z^n}$.

3. Originally Posted by chiph588@
What about the function $\displaystyle \displaystyle f(z)=e^{1/z}-\frac1z = 1+\sum_{n=2}^{\infty} \frac{1}{n!z^n}$.
Oh, that would be a counter example?
I realize that $\displaystyle e^{1/z}=\sum _{n=0}^{+\infty} \frac{1}{z^n n!}=1+ \sum _{n=1}^{+\infty} \frac{1}{z^n n!}$.
But I'm not sure how to reach $\displaystyle e^{1/z}-\frac1z = 1+\sum_{n=2}^\infty \frac{1}{n!z^n}$.
Also, $\displaystyle Res(e^{1/z},z=0)=1$ while $\displaystyle Res(-1/z,z=0)=-1$ so indeed $\displaystyle Res(f,z=0)=0$. So the singularity of $\displaystyle e^{1/z}$ is essential (there's an infinity of negative powers of z in the series expansion of the function) and so should be the one of f... So that after all, z isn't bounded nor analytic in a neighborhood of $\displaystyle z=0$?
Should I use Cauchy-Riemann equations to show that f isn't analytic? How would you do it?

4. Originally Posted by arbolis
Oh, that would be a counter example?
I realize that $\displaystyle e^{1/z}=\sum _{n=0}^{+\infty} \frac{1}{z^n n!}=1+ \sum _{n=1}^{+\infty} \frac{1}{z^n n!}$.
But I'm not sure how to reach $\displaystyle e^{1/z}-\frac1z = 1+\sum_{n=2}^\infty \frac{1}{n!z^n}$.
Also, $\displaystyle Res(e^{1/z},z=0)=1$ while $\displaystyle Res(-1/z,z=0)=-1$ so indeed $\displaystyle Res(f,z=0)=0$. So the singularity of $\displaystyle e^{1/z}$ is essential (there's an infinity of negative powers of z in the series expansion of the function) and so should be the one of f... So that after all, z isn't bounded nor analytic in a neighborhood of $\displaystyle z=0$?
Should I use Cauchy-Riemann equations to show that f isn't analytic? How would you do it?
$\displaystyle \displaystyle e^{1/z} = \sum_{n=0}^\infty \frac{1}{n!z^n} = 1+\frac1z+\sum_{n=2}^\infty \frac{1}{n!z^n}$

Therefore $\displaystyle \displaystyle e^{1/z}-\frac1z = 1+\sum_{n=2}^\infty \frac{1}{n!z^n}$

5. Or just look at $\displaystyle 1/z^2$... it has a vanishing residue at $\displaystyle 0$.

6. Originally Posted by Bruno J.
Or just look at $\displaystyle 1/z^2$... it has a vanishing residue at $\displaystyle 0$.
Ok thank you. And to show it's not analytic, I could either show that f'(z) doesn't exist or show that C-R equations aren't satisfied?

7. Originally Posted by Bruno J.
Or just look at $\displaystyle 1/z^2$... it has a vanishing residue at $\displaystyle 0$.
But that's a removable singularity...

8. Originally Posted by chiph588@
But that's a removable singularity...
And I think it's analytic in a neiborhood of 0. Is that right? I've shown that $\displaystyle f'(z)=\frac{-2}{z^3}$ for any $\displaystyle z \neq 0$.

9. Originally Posted by chiph588@
But that's a removable singularity...
No, it's not! It's a second-order pole.

10. Originally Posted by arbolis
And I think it's analytic in a neiborhood of 0. Is that right? I've shown that $\displaystyle f'(z)=\frac{-2}{z^3}$ for any $\displaystyle z \neq 0$.
No, it's not analytic at the origin. It's analytic in a punctured neighbourhood of the origin, but not at the origin. (Unless you consider poles to be regular points, which is sometimes done.)

11. Whoops! I forgot my definitions!

I thought a removable singularity was when you could multiply by a polynomial of an arbitrary degree to remove the singularity.

12. Originally Posted by chiph588@
Whoops! I forgot my definitions!

I thought a removable singularity was when you could multiply by a polynomial of an arbitrary degree to remove the singularity.
No worries!
What you describe is a pole. A removable singularity is just camouflaged regular point!

13. Originally Posted by Bruno J.
No, it's not analytic at the origin. It's analytic in a punctured neighbourhood of the origin, but not at the origin. (Unless you consider poles to be regular points, which is sometimes done.)
Ah I see. I'm not sure it would fit as a counter example due to the "punctured" neighborhood.
By the way, last question before my exam: if $\displaystyle f(z)=g(z)+h(z)$, does $\displaystyle Res(f,z_0)=Res(g,z_0)+Res(h,z_0)$ assuming that f (and maybe g and h) has a pole at $\displaystyle z_0$.

14. Yup, by linearity of the integral!