Complex analysis - finding the order of poles

**artvandalay11** · May 9th 2009, 02:15 PM

I am stuck on the following question:

Find the poles of z/(1-exp(z)) and determine their orders.

I have the poles as being all z*2*Pi*i*k with k an integer other than 0 as I am not sure what happens at 0.

How do I find the order though? It seems to me that writing the series expansion out does not help much.

Thanks for any and all help (my guess is that the order is 1)

**TheEmptySet** · May 9th 2009, 03:31 PM

Originally Posted by artvandalay11

I am stuck on the following question:

Find the poles of z/(1-exp(z)) and determine their orders.

I have the poles as being all z*2*Pi*i*k with k an integer other than 0 as I am not sure what happens at 0.

How do I find the order though? It seems to me that writing the series expansion out does not help much.

Thanks for any and all help (my guess is that the order is 1)

The pole is of order zero it is a removable singularity at 0

Remember that

$e^{z}=1+z+\frac{z^2}{2}+\frac{z^3}{6}+....$

So

$1-e^{z}=-(z+\frac{z^2}{2}+\frac{z^3}{6}+....)$

Now we have

$\frac{z}{1-e^z}=\frac{-z}{(z+\frac{z^2}{2}+\frac{z^3}{6}+....)}=$

$\frac{-1}{(1+\frac{z}{2}+\frac{z^2}{6}+....)}$

Now by long division we get

$\frac{-1}{(1+\frac{z}{2}+\frac{z^2}{6}+....)}=-1+\frac{z}{2}-\frac{1}{12}z^2+\frac{1}{720}z^4-...$

**artvandalay11** · May 9th 2009, 03:47 PM

thank you for confirming my suspicions... I know I said I thought the pole had order 1, but I also have exactly what you wrote on my paper, I just was not sure if it was removable because calculating the limit as a test for removable singularities seemed to be difficult... again thanks a lot

**artvandalay11** · May 10th 2009, 04:59 PM

I was thinking about this some more, and the function should have a pole at $z=2\pi\imath$ since $\lim_{z \to 2\pi\imath}|f(z)|=\infty$ (also at any multiple of $z=2\pi\imath$ )

The series expansion is correct, but the function still tends to infinity. To show that $z=2\pi\imath$ is a pole we have to be able to show $\lim_{z \to 2\pi\imath}(z-2\pi\imath)f(z)=0$

I can't seem to be able to do that

**ThePerfectHacker** · May 10th 2009, 05:53 PM

Originally Posted by artvandalay11

I was thinking about this some more, and the function should have a pole at $z=2\pi\imath$ since $\lim_{z \to 2\pi\imath}|f(z)|=\infty$ (also at any multiple of $z=2\pi\imath$ )

The series expansion is correct, but the function still tends to infinity. To show that $z=2\pi\imath$ is a pole we have to be able to show $\lim_{z \to 2\pi\imath}(z-2\pi\imath)f(z)=0$

I can't seem to be able to do that

Write, $\frac{z}{1-e^z} = \frac{f(z)}{g(z)}$ . If $g'(z_0) \not = 0$ then the residue of the pole at $z_0$ is $\frac{f(z_0)}{g'(z_0)} = -\frac{z_0}{e^{z_0}}$ .
Thus, we see that the residues are $-2\pi i k$ for $k\in \mathbb{Z}$ .

**artvandalay11** · May 10th 2009, 06:17 PM

Alright, but I need to find the order of the poles

**ThePerfectHacker** · May 10th 2009, 07:30 PM

Originally Posted by artvandalay11

Alright, but I need to find the order of the poles

That theorem only works if the order is one!
(For k=0 we get residue 0 so there is no pole).

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