We'll show you two way of calculation of the Residue in consideration.

$$f(z) = \frac{z\sin(z)}{1-\cos(z)}$$

I'm interested to calculate the residues in $2\pi$ and $-2\pi$. I choose one of them.($2\pi$)

After checked the type of singularity(simple pole), I choose one of two manner to get the residue.

(1) $$\lim_{z\rightarrow 2\pi} \frac{z\sin(z)}{D(1-\cos(z))} = \lim_{z\rightarrow 2\pi} \frac{z\sin(z)}{\sin(z)} = \lim_{z\rightarrow 2\pi} z = 2\pi$$

(2) $$\lim_{z\rightarrow 2\pi} \frac{z\sin(z)}{1-\cos(z)} = \lim_{z\rightarrow 2\pi} \frac{z\sin(z-2pi)(z-2\pi)^2}{(1-\cos(z))(z-2\pi)} = 4\pi$$

Which of two I have to choose?

Thanks for the answers

  • $\begingroup$ What is your $D$? $\endgroup$ – user87543 Jan 16 '14 at 10:47
  • $\begingroup$ It's the derivative of the argument. $\endgroup$ – Pixel Jan 16 '14 at 10:48
  • $\begingroup$ Yes, It's derivative of the argument. Sorry, I don't know very well Latex. $\endgroup$ – GenKs Jan 16 '14 at 10:49
  • $\begingroup$ I am not sure If $\lim_{z\rightarrow 2\pi} \frac{zsin(z)}{1-cos(z)}$ gives you the residue.. $\endgroup$ – user87543 Jan 16 '14 at 10:52
  • $\begingroup$ @GenKs : Please look at Mr.Daniel's answer... the pole is not simple pole so your first procedure does not work... $\endgroup$ – user87543 Jan 16 '14 at 11:00

Here, the denominator has a double zero, which is partially compensated by a zero of the numerator to produce a simple pole altogether, thus the residue is not given by the formula

$$\operatorname{Res} \left(\frac{f(z)}{g(z)}; z_0\right) = \frac{f(z_0)}{g'(z_0)}$$

that one uses for simple zeros of the denominator.

If we write $f(z) = (z-z_0)\cdot f_1(z)$ with $f_1(z_0) = f'(z_0) \neq 0$, and $g(z) = (z-z_0)^2\cdot g_2(z)$ with $g_2(z_0) = \frac12 g''(z_0) \neq 0$, we see that the residue of $\frac{f}{g}$ in $z_0$ is

$$\operatorname{Res} \left(\frac{f(z)}{g(z)}; z_0\right) = \operatorname{Res} \left(\frac{(z-z_0)f_1(z)}{(z-z_0)^2g_2(z)}; z_0\right) = \frac{f_1(z_0)}{g_2(z_0)} = 2\frac{f'(z_0)}{g''(z_0)}.$$

That means your second way is correct.

More generally, if the denominator has a zero of order $n$, and the numerator a zero of order $n-1$ in $z_0$, the Taylor series of numerator and denominator start

$$\begin{align} f(z) &= \frac{f^{(n-1)}(z_0)}{(n-1)!}(z-z_0)^{n-1} + \dotsc,\\ g(z) &= \frac{g^{(n)}(z_0)}{n!}(z-z_0)^n + \dotsc, \end{align}$$

and the residue of $f/g$ in $z_0$ is then

$$\operatorname{Res}\left(\frac{f(z)}{g(z)}; z_0\right) = \frac{f^{(n-1)}(z_0)/(n-1)!}{g^{(n)}(z_0)/n!} = n\cdot \frac{f^{(n-1)}(z_0)}{g^{(n)}(z_0)}.$$

  • $\begingroup$ Thank you very much, Mr. Fischer. Just a last doubt: what's the reason of why we've chose the terms of taylor series of numerator and denominator of first and second order, respectively. is It associeted with the zero's order? $\endgroup$ – GenKs Jan 16 '14 at 11:29
  • $\begingroup$ Ah, I forget.. why is It partially compensated? How can we state this? $\endgroup$ – GenKs Jan 16 '14 at 11:32
  • $\begingroup$ It's because here, the numerator has a simple zero, and the denominator a double zero. If the zero of the numerator $f$ has order $k$, and the zero of the denominator $g$ order $k+1$, then $\frac{f}{g}$ has a simple pole, and the residue is the quotient of the $k$-th Taylor coefficient of $f$ and the $k+1$-th of $g$, which is $(k+1)\dfrac{f^{(k)}(z_0)}{g^{(k+1)}(z_0)}$. If the order of the zero of $g$ is more than one larger than the order of the zero of $f$, it gets more complicated. $\endgroup$ – Daniel Fischer Jan 16 '14 at 11:35
  • $\begingroup$ Ok, now I understand much more. Mr. Fischer, thank you again for your clarification. $\endgroup$ – GenKs Jan 16 '14 at 11:41

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