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I have try to use the equation $$ Res(f;z_0)=\lim_{z\to z_0}\frac1{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)] $$ But very soon I stuck, is that a good way to solve it? enter image description here

user111605
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1 Answers1

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A few hints:

1) Since you have a double pole at $z=z_0$, your $f$ has the form

$$f(z) = \frac{p(z)}{(z-z_0)^2 r(z)}$$

i.e., $q(z) = (z-z_0)^2 r(z)$, where $r$ is analytic at $z=z_0$.

2) Use the given expression for the residue with $m=2$. This means you will be taking the derivative of $p(z)/r(z)$.

3) Finally, you need things in terms of $q$, not $r$, you need to evaluate the second and third derivatives of $q(z)=(z-z_0)^2 r(z)$ at $z=z_0$. This will give you expressions for $r(z_0)$ and $r'(z_0)$ as needed if you did everything right in step 2).

Ron Gordon
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  • After I find out$$ r(z_0) $$and $$r'(z_0) $$and sub into the formula, it still far away from the question $$ \frac{P'(z)(z-z_0)^3}{Q'(z)(z-z_0)-2Q(z)} $$ Is there something goes wrong? – user111605 Nov 26 '13 at 17:42
  • You have to plug in $z=z_0$ into the derivative formula. For example, $$q'(z) = 2 (z-z_0) r(z) + (z-z_0)^2 r'(z)$$ so that $q'(z_0)=0$. What about the second and third derivatives of $q$? – Ron Gordon Nov 26 '13 at 17:47
  • $$ Q"(z_0)=2r(z_0), Q"'(z_0)=2r'(z_0) $$ But how can a single term,the derivative of P(z)/r(z), expand to two terms? It is still quite far from the question. – user111605 Nov 26 '13 at 18:54
  • @user111605: do you know the quotient rule? I hope so if you are studying this stuff. – Ron Gordon Nov 26 '13 at 19:00