# Math Help - Complex Analysis singularities question

1. ## Complex Analysis singularities question

question:
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locate each of the isolated singularities of the given function and tell whether it is removable singularity, a pole, or an essential singularity. If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole.

(e^z-1)/(e^2z-1)

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What I first tried to do was factoring the bottom.

(e^z-1)/((e^z+1)(e^z-1))

but realized

e^z-1 =0 for z = i(2(pi)n) n = 0, 1 , 2 ...
e^z+1=0 for z= i((pi)+2(pi)n) n = 0, 1, 2...

so basically, denominator = 0 whenever z = i(pi)n , n = 0, 1, 2....

I tried to make this into An(z-z0)^n , but was unsuccessful.

But can't really tell if this really is a removable singularity at z = i(pi)n, n= 0, 1, 2.... (I think |f(z)| remains bounded as z --> any of those points, so it is removable?)

I thought about essential singularities, but I don't think that's the case...

If anyone can solve this problem/ explain, that would be of great help. Thanks.

2. Originally Posted by hohoho00
question:
------------------------------------------

locate each of the isolated singularities of the given function and tell whether it is removable singularity, a pole, or an essential singularity. If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole.

(e^z-1)/(e^2z-1)

------------------------------------------

What I first tried to do was factoring the bottom.

(e^z-1)/((e^z+1)(e^z-1))

but realized

e^z-1 =0 for z = i(2(pi)n) n = 0, 1 , 2 ...
e^z+1=0 for z= i((pi)+2(pi)n) n = 0, 1, 2...

so basically, denominator = 0 whenever z = i(pi)n , n = 0, 1, 2....

I tried to make this into An(z-z0)^n , but was unsuccessful.

But can't really tell if this really is a removable singularity at z = i(pi)n, n= 0, 1, 2.... (I think |f(z)| remains bounded as z --> any of those points, so it is removable?)

I thought about essential singularities, but I don't think that's the case...

If anyone can solve this problem/ explain, that would be of great help. Thanks.
Removable singularities at z such that $e^z - 1 = 0$ (since the limit exists and is finite).

Essential singularities at z such that $e^z + 1 = 0$ (since the Laurent series doesn't terminate).

3. Thank you ^^

I was wondering though, after removing (e^z -1), we get the equation:

1/ (e^z+1)

but is it possible to express this in laurent series? Or can we still say it is a laurent series despite all the series is at the denominator?

4. Originally Posted by hohoho00
Thank you ^^

I was wondering though, after removing (e^z -1), we get the equation:

1/ (e^z+1)

but is it possible to express this in laurent series? Or can we still say it is a laurent series despite all the series is at the denominator?
Yes, this function does have a Laurent series in the neighbourhood of any singularity. In fact, the singularities are poles of order 1, because the denominator has zeros of order 1.

To see that there is a simple pole at $z-i\pi$, for example, you can use l'Hôpital's rule to check that $\lim_{z\to i\pi}\frac{z-i\pi}{e^z+1} = \frac1{e^{i\pi}}=-1$.

5. Originally Posted by Opalg
Yes, this function does have a Laurent series in the neighbourhood of any singularity. In fact, the singularities are poles of order 1, because the denominator has zeros of order 1.

To see that there is a simple pole at $z-i\pi$, for example, you can use l'Hôpital's rule to check that $\lim_{z\to i\pi}\frac{z-i\pi}{e^z+1} = \frac1{e^{i\pi}}=-1$.
My mistake. Thanks for the catch.

6. Hmm.... does that mean it just has a simple pole at all i(pi)n? and there's no essential singularities?

7. Originally Posted by hohoho00
Hmm.... does that mean it just has a simple pole at all i(pi)n? and there's no essential singularities?
My chance at redemption.

$e^z + 1 = 0 \Rightarrow z = (\pi + 2n\pi)i = \pi (2n+1) i$ where n is an integer.

Now apply the same argument that Opalg has already given to show that these are simple poles.