# Calculating residue of pole of order $2$

Is there a good way to compute the residue of $f(z)=\dfrac{1+z}{1-\sin z}$ at $z=\pi/2$, which is a pole of order $2$?

Using the residue calculation formula yields

$$\text{Res}_{z=\pi/2}f(z)=\lim_{z\rightarrow\pi/2}\dfrac{d}{dz}\left(\left(z-\dfrac\pi2\right)^2f(z)\right)$$

The derivative is quite ugly, and calculating the limit requires L'Hospital probably twice (or more). The calculation is just too much. Is there a better way?

• Have you tried expanding $\sin z$ in a Taylor series? Then you are left with the residue where you have $\frac{1}{z}$. Nov 5, 2013 at 4:21
• @JeremyUpsal I'm not quite sure what you mean there. Could you show with a formula? Nov 5, 2013 at 4:25
• Hint: The residue of $f(z)$ at $z=a$ is the coefficient of $(z-a)^{-1}$ in the Laurent series of $f(z)$ centered at $z=a$. Nov 5, 2013 at 4:38
• In other words, if $f(z) = \sum_{k=-\infty}^{\infty} \beta_k (z-a)^k$, then $\operatorname{Res}(f,a) = \beta_{-1}$. Nov 5, 2013 at 4:49
• Is that really any less complicated do find? Nov 5, 2013 at 4:56

Consider the residue of $f(z)/g(z)$ at the double pole $z=a$. Because $a$ is a double zero of $g(z)$, write

$$g(z) = (z-a)^2 p(z)$$

where $p(a) \ne 0$ and is analytic, etc. etc.

Then

$$\operatorname*{Res}_{z=a} \frac{f(z)}{g(z)} = \left [\frac{d}{dz} \frac{f(z)}{p(z)} \right ]_{z=a}$$

Now,

$$\frac{d}{dz} \frac{f(z)}{p(z)} = \frac{f'(z) p(z)-f(z) p'(z)}{p(z)^2}$$

Also, note that

$$g(z) = \frac12 g''(a) (z-a)^2 + \frac16 g'''(a) (z-a)^3+\cdots = p(a) (z-a)^2 + p'(a) (z-a)^3+\cdots$$

Therefore

$$p(a) = \frac12 g''(a)$$

and

$$p'(a) = \frac16 g'''(a)$$

Thus

$$\operatorname*{Res}_{z=a} \frac{f(z)}{g(z)} = \frac{6 f'(a) g''(a) - 2 f(a) g'''(a)}{3 [g''(a)]^2}$$

In your case, $f(z)=1+z$ and $g(z)=1-\sin{z}$. The residue at $z=\pi/2$ is then $2$.

I love using Taylor series expansions, as I find they're often the easiest way by hand. Let $$z=u+\pi/2$$, then $$f(z)=\frac{1+u+\frac{\pi}{2}}{1-\sin(u+\frac{\pi}{2})}=\frac{1+\frac{\pi}{2}+u}{1-\cos u}=\frac{1+\frac{\pi}{2}+u}{1-(1-u^2/2!+u^4/4!-\cdots)}.$$ Simplifying the denominator and extracting a factor of $$u^2$$, $$\frac{1}{u^2}\frac{1+\frac{\pi}{2}+u}{1/2!-u^2/4!+u^4/6!-\cdots}=\frac{2}{u^2}\frac{1+\frac{\pi}{2}+u}{1-2u^2/4!+2u^4/6!-\cdots}.$$ Now using $$\frac{1}{1-z}=1+O(z)$$, we get $$\frac{2}{u^2}(1+\frac{\pi}{2}+u)(1+O(u^2))=\frac{2}{u^2}(1+\frac{\pi}{2})+\frac{2}{u}+O(1),$$ or in other words $$f(z)=\frac{2(1+\frac{\pi}{2})}{(z-\frac{\pi}{2})^2}+\frac{2}{z-\frac{\pi}{2}}+O(1),$$ so clearly $$\text{Res}(f,\frac{\pi}{2})=2$$.

• Why does $\frac{1}{1-z} = 1 + O(z)$ hold? Jan 3, 2023 at 14:23
• @Dedekind because $\frac{1}{1-z}=1+z+z^2+z^3\cdots=1+O(z)$ where $+O(z)$ means "plus on the order of $z$" which loosely means all the additive terms including $z$, $z^2$, $z^3$, etc. We use this notation because we don't care about those terms when computing the residue since the residue comes from the coefficient of the $\frac{1}{z-a}$ term. Jan 3, 2023 at 17:27

My other answer (and Antonios above) requires knowledge of finding poles through Laurent series, which I assumed you have done (I hope!), but it is possible you have not. Instead, you can use that when $f(z) = \frac{p(z)}{q(z)}$ has a pole at $z=z_0$ where $p(z)$ and $q(z)$ are analytic in any neighborhood of $z_0$, you have $Res(f;z=z_0) = \frac{p(z_0)}{q'(z_0)}$. If, as in this case, $q'(z_0)=0$, you have to change this formula. It was derived from the Taylor series of $p(z)$ and $q(z)$ about $z=z_0$. So, we can take the next order term, namely $Res(f,z=z_0) = \frac{p'(z_0)}{q''{z_0}/2!}$. Here you have that case and can apply this formula nicely, but it is important to see where it comes from which is the residue from the Laurent series as described before.

• Should I give the derivation for something like this? Or is it okay to post this result without a derivation. I am relatively new to the site, it would be great to know what is preferred by others. Nov 5, 2013 at 5:02
• I think the general mantra would be "more is always good, up to the point where you get that icky 'I just did this guy's homework for him' feeling". Nov 5, 2013 at 5:24
• Okay good. I tried to keep it general but I omitted much of the proof although it is easy to reproduce if you have any experience with residues. Nov 5, 2013 at 5:37