This is the general formula often used to calculate the residue of a complex function around a pole. (Taken from Wikipedia)

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But say if I have the function $$f(z) = \frac{z^4}{z^6+1}$$

it has a pole of order $n =1$ at $z_0 = e^{\tfrac{i\pi}{6}}$. Then to calculate its residue at $z_0$ by the above formula I'd have

$$\operatorname{Res}(f, z_0) = \lim_{z \to z_0} (z-z_0) f(z) = \lim_{z \to z_0} (z-z_0) \frac{z^4}{z^6+1} = \lim_{z \to z_0} \frac{z^5-z_0z^4}{z^6 + 1} = 0$$

because $$\lim_{z \to z_0} z^5 - z_0z^4 = z_0^5 - z_0^5 = 0$$

However my lecturer calculated the residue of $f$ at $z_0$ as follows

$$\operatorname{Res}(f, z_0) = \left[\frac{z^4}{6z^5}\right]_{z = z_0} = \frac{1}{6z_0} = \frac{1}{6e^{\tfrac{i \pi}{6}}}$$

He's basically taken the derivative of the denominator and evaluated this new function, with the old numerator and differentiated denominator, at $z= z_0$. How exactly could this be done? I don't see how the formula from Wikipedia implies that one can do this to find residues.

Also am I making an error in evaluating the residues above?

  • $\begingroup$ Your computation is incorrect, because $z_0^6+1=0$ in the denominator. $\endgroup$ – Lorenzo Quarisa Jun 1 '18 at 13:03
  • $\begingroup$ So since $z_0$ is a root of $z^6+1$, you can factor out $(z-z_0)$ in $z^6+1$, you will get $z^6+1=(z-z_0)p(z)$ for some polynomial $p$ and then the limit for $z\to z_0$ becomes $z_0^4/p(z_0)$. $\endgroup$ – Lorenzo Quarisa Jun 1 '18 at 13:06
  • $\begingroup$ @LorenzoQ. Ohh I see applying L'Hopitals rule I'll end up with the way in which my lecturer calculated the residue. $\endgroup$ – Perturbative Jun 1 '18 at 13:07
  • $\begingroup$ Sure, that works as well. $\endgroup$ – Lorenzo Quarisa Jun 1 '18 at 13:11

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