I have the following contour integral (with C the positively oriented unit circle centered at the origin):

$$ \frac{-i}{4}\int_{C}\frac{\left(z^2+1\right)^2}{z\left(-z^4+3z^2-1\right)}dz $$

It has isolated singularities inside $C$ at $z = 0, \pm\sqrt{\frac{3-\sqrt{5}}{2}}$

$z = 0$ is a simple pole and the residue at that point is easily computed. I am unsure how to compute the residues at $z = \pm\sqrt{\frac{3-\sqrt{5}}{2}}$, without having to write the Laurent series?


If $z_0$ is a pole of $f$ of order $m$, you can use the formula $$Res(f,z_0)=\frac{1}{(m-1)!} \lim_{z\to z_0} [(z-z_0)^m \cdot f(z)]^{(m-1)}$$

where $g^{(n)}$ means the $n$-th derivative of $g$.

Please ask if you need work on how that formula is obtained.

  • $\begingroup$ sorry, I can't see the difference actually, $f^{(n)}$ means the $n$-th derivative. I'll explain it anyway, thanks. $\endgroup$ Jan 9 '17 at 22:39
  • $\begingroup$ I am familiar with that formula, however I don't see a way to factor the denominator in such a way that the order is the power of $(z-z_0)$. (I might be missing something obvious) $\endgroup$
    – user405561
    Jan 9 '17 at 22:44
  • $\begingroup$ @Tom The notation $f^{(n)}$ is pretty standard for the $n$th derivative of $f$. $\endgroup$ Jan 9 '17 at 22:44
  • $\begingroup$ @user405561 you can solve the equation $-z^4+3z^2-1$ by substituting $t=z^2$. $\endgroup$ Jan 9 '17 at 22:47
  • $\begingroup$ Indeed, that's how I arrived at $$z0 = \pm\srqt{\frac{3-\sqrt{5}}{2}}$$, I still don't know $m$, though. $\endgroup$
    – user405561
    Jan 9 '17 at 22:51

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.