# Find and classify the isolated singularity

1. May 4, 2017

### fishturtle1

1. The problem statement, all variables and given/known data
Find and classify the singularity.

f(z) = sin(z) / z

2. Relevant equations
sin(z) = (z - z3 / 3! + z5 / 5! - z7 / 7! + z9 / 9! - ...

3. The attempt at a solution
so the singularity is at z = 0.

f(z) = (1/z) * (z - z3 / 3! + z5 / 5! - z7 / 7! + z9 / 9! - ...)

f(z) = 1 - z2 / 3! + z4 / 5! - z6 / 7! + z8 / 9! ...

and these are terms starting at index 0, because the series for sin(z) is Σ(-1)k * (z2k + 1 / (2k+1)!), for k >= 0.

The definition of removable singularity is if cn = 0 for n = -1, -2, -3 ... then f(z) has a removable singularity at α.

ANSWER: So the singularity z = 0 is removable because the series for sin(z) starts at k = 0, and so any term k<0 must have coefficient 0.

i'm really confused how to check whether the singuarlity is removable, a pole, or essential. The definition doesn't make sense to me. How can I check if ck = 0 for all k < 0, if there are no terms for k < 0 in the power series for sin(z)?

2. May 4, 2017

### Ray Vickson

Why do you think you need to check? You just said that there are no terms for k < 0 in the power series.

3. May 4, 2017

### fishturtle1

yeah... so "no terms for k < 0" <=> "ck = 0 for all k < 0"?

4. May 4, 2017

### Ray Vickson

What do YOU think?

5. May 5, 2017

### fishturtle1

Ok, so in order to find out what kind of singularity it is, I need to write this as a Laurent series.

So a Laurent series, as I understand it, is f(z) rewritten as
Σ Ck(z-z0)k, for k >= 0 + Σ Ck(z - z0)k, for k < 0.

I've got the first series.. (1 - z2/3! + z4/5! - z7/8! + ...

but I need to see the values for the negative indexes.. so I rewrite the series like this

$\frac{1}{z} * \sum_{n=1}^{-\infty} (-1)^{-k-1} \frac{z^{-2k-1}}{|z|!} =\frac{1}{z} * ( \frac{1}{z^{3}3!} - \frac{1}{z^{5}5!} + \frac{1}{z^{7}7!} - ...) = \frac{1}{z^{4}3!} - \frac{1}{z^{6}5!} + \frac{1}{z^{8}7!} - ...$

Looking at this.. I think that these negative index value have coefficients of $\frac{1}{|k|!}$ which isn't 0 which is what I expected

6. May 5, 2017

### Ray Vickson

An ordinary power series $\sum_{n \geq 0} c_n z^n$ IS a Laurent series with all negative powers absent. Usually, though, we do not refer to it as a Laurent series (with no negative-power terms); it is easier to just call it a power series.

7. May 6, 2017

### fishturtle1

But if there are no terms raised to z raised to a negative exponent, then how can there exist nonexistent terms that have Ck = 0?

If a power series is a series that has all negative powers absent, then how can there be terms:
C-1 (z - z0) + C-2 (z - z0) + .... = 0, if the series starts at k=0?

I know 0 + Σck(z-z0) = Σck(z-z0) how do we know Ck = 0.. what if (z-z0) = 0 and Ck is some number?

8. May 6, 2017

### fishturtle1

sorry for double post

9. May 6, 2017

### Ray Vickson

It's easy: $0 (z-z_0)^{-1} + 0 (z-z_0)^{-2} + 0 (z-z_0)^{-3} + \cdots$ is just zero. I genuinely cannot understand why you have a problem with that, but at this point I am quitting this thread.