I know that, since the complex function $$f(z)=\frac{1}{z(e^z-1)}$$ have a pole of 2nd order in z=0, I should be able to represent it as: $$f(z)=\frac{g(z)}{(z-0)^2}$$ Being g(z) a function that is both analytic and different from zero when z=0.

But the only expression for g(z) that I was able to get to was $$g(z)=\frac{z}{e^z-1}$$ wich doesn't meets the criteria explained before.

Can anyone say what am I getting wrong? Does f(z) really have a pole of 2nd order in zero?

  • $\begingroup$ "wich doesn't meets the criteria explained before" Hmmm... why do you think it does not? $\endgroup$ – Did Dec 10 '18 at 2:50
  • $\begingroup$ @Did well, because g(z) has a singularity when z=0, doensn´t it? That would make it a non-analytic function in this point from what I know, but I could be wrong of course $\endgroup$ – Rodrigo Castañon Dec 10 '18 at 10:37
  • $\begingroup$ $g(z)$ has a removable singularity at $z=0$, since $\lim_{z\to 0} g(z) = 1$. You can make it analytic by defining $$ g(z) = \begin{cases} \dfrac{z}{e^z-1}, & z \ne 0 \\ 1, & z = 0 \end{cases} $$ $\endgroup$ – Dylan Dec 10 '18 at 14:42
  • $\begingroup$ How is the post below that you chose to accept, answering the question? $\endgroup$ – Did Dec 10 '18 at 15:16
  • $\begingroup$ @Dylan yes, I realized that after a while, thank you! $\endgroup$ – Rodrigo Castañon Dec 10 '18 at 17:51

I'm just taking my first complex analysis but from my understanding you are correct in the fact that pole z= 0 is of second order.

Using the formula for higher order poles.

$$Res_{z \rightarrow x} = \frac{1}{(n-1)!} lim_{z\rightarrow x} \frac{d^{n-1}}{dz^{n-1}}(z-x)^n f(z)$$

Where x is the pole 0 and n is the order.

Using this formula:

$$Res_{z \rightarrow 0} = \frac{1}{(2-1)!} lim_{z\rightarrow 0} \frac{d^{2-1}}{dz^{2-1}}(z-0)^2 f(z)$$

$$Res_{z \rightarrow 0} = \frac{1}{(1)!} lim_{z\rightarrow x} \frac{d}{dz}(z-0)^2 \frac{1}{z(e^z - 1)}$$

$$ lim_{z\rightarrow 0} \frac{d}{dz}(z)^2 \frac{1}{z(e^z - 1)}$$ $$ lim_{z\rightarrow 0} \frac{d}{dz}\frac{z }{(e^z - 1)}$$ $$ lim_{z\rightarrow 0} \frac{(e^z - 1) - (e^z)z}{(e^z - 1)^2}$$

Now just do a couple L'Hopitals and you should get: Res at z=0: $$= - \frac{1}{2}$$

  • 1
    $\begingroup$ Thank you! I've just started to study the subject too and your thoughts on the question were very helpful :) $\endgroup$ – Rodrigo Castañon Dec 10 '18 at 11:06
  • $\begingroup$ Glad to be of help! @RodrigoCastañon $\endgroup$ – Safder Dec 10 '18 at 22:32

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