6
$\begingroup$

I'm trying to compute the residue $\displaystyle\operatorname{Res}\left(\frac{1}{(z^2+1)^7},i\right)$.

I know that there is the formula:

$$\operatorname{Res}(f,z_0)=\frac{1}{(m-1)!}\lim_{z\rightarrow z_0 }[(z-z_0)^mf(z)]^{(m-1)}$$

for a pole with order $m$.

But I'm pretty sure that I should not try to compute the 6th derivative of $\dfrac{1}{(z+i)^7}$.

Is there another way to compute the residue beside this formula?

$\endgroup$

2 Answers 2

16
$\begingroup$

Hmm. \begin{align*} \lim_{z\to i}\left[\frac{1}{(z+i)^7}\right]^{(6)}&= \lim_{z\to i}\left[(z+i)^{-7}\right]^{(6)} \\ &=\lim_{z\to i}\left[-7(z+i)^{-8}\right]^{(5)} \\ &=\lim_{z\to i}\left[(-7)(-8)(z+i)^{-9}\right]^{(4)} \dots \end{align*} Calculating the derivatives doesn't seem too bad.

$\endgroup$
10
$\begingroup$

Expanding at $i$, let $w = z - i$:

$$\begin{array}{rcl} \operatorname{Res}(f,z_0) &=& \operatorname{Res}\left(\dfrac1{w^7 (w+2i)^7},0\right) \\ &=& \operatorname{Res}\left(\dfrac1{w^7}(w+2i)^{-7},0\right) \\ &=& \operatorname{Res}\left(\dfrac1{w^7}(2i+w)^{-7},0\right) \\ &=& (2i)^{-7} \operatorname{Res}\left(\dfrac1{w^7}(1-0.5iw)^{-7},0\right) \\ &=& (2i)^{-7} \displaystyle \operatorname{Res}\left(\dfrac1{w^7} \sum_{n=0}^\infty \binom {-7} n (-0.5iw)^n,0\right) \\ &=& \displaystyle (2i)^{-7} \binom {-7} 6 (-0.5i)^6 \\ &=& -\dfrac{231}{2048}i \end{array}$$

$\endgroup$
6
  • $\begingroup$ How did $(1-0.5iw)^{-7}$ turn into a sum? $\endgroup$
    – bp7070
    May 30, 2018 at 15:47
  • 3
    $\begingroup$ Generalized binomial formula $\endgroup$
    – Kenny Lau
    May 30, 2018 at 15:48
  • $\begingroup$ OK thank you. I'm not realy sure that this is raleted but is it possible to expend the function to Laurent series and take the $c_{-6}$ coefficiant? Would it be easier? $\endgroup$
    – bp7070
    May 30, 2018 at 15:52
  • 2
    $\begingroup$ That's the same $\endgroup$
    – Kenny Lau
    May 30, 2018 at 15:52
  • 2
    $\begingroup$ Extracting the $w^{-1}$ coefficient in the expression inside Res, which is the $w^6$ coefficient inside the summation. $\endgroup$
    – Kenny Lau
    May 30, 2018 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.