Residues of $ \frac{z^4}{1+z^6} $

Question

I am trying to compute all 6 the residues of $ \frac{z^4}{1+z^6} $. I tried the straightforward way first of finding all the points where the denominator is 0 etc but it became way too complicated. Any ideas?

Answer 1

Let $a$ be a pole of $f(z)=z^4/(z^6+1)$. Then $a^6=-1$. Also $a$ is a simple pole. The residue is thus $$\lim_{z\to a}(z-a)f(z)=\lim_{z\to a}\frac{z^4(z-a)}{z^6-a^6} =\frac{a^4}{\lim_{z\to a}\frac{z^6-a^6}{z-a}}.$$ But $$\lim_{z\to a}\frac{z^6-a^6}{z-a}=g'(a)$$ when $g(z)=z^6$ etc.

Answer 2

$z^6 + 1$ has roots $e^{\frac {(2k+1)\pi }{6}i}$

We might factor the denominator like this.

$(z-e^{\frac {\pi}{6} i})(z^5 + z^4e^{\frac {\pi}{6}i} + z^3e^{\frac {2\pi}{6}i} + z^2e^{\frac {3\pi}{6}i} + ze^{\frac {4\pi}{6}i} + e^{\frac {5\pi}{6}i})$

$\lim_\limits{z\to e^{\frac {\pi}{6}i}} \frac {z^4}{z^6+1} (z-e^{\frac {\pi}{6}i}) = \lim_\limits{z\to e^{\frac {\pi}{6}i}} \frac {z^4}{(z^5 + z^4e^{\frac {\pi}{6}i} + z^3e^{\frac {2\pi}{6}i} + z^2e^{\frac {3\pi}{6}i} + ze^{\frac {4\pi}{6}i} + e^{\frac {5\pi}{6}i})}$

As $z$ approaches $e^{\frac {pi}{6} i},$ each term in the denominator approaches $e^{\frac {5\pi}{6}i}$

$\lim_\limits{z\to e^{\frac {\pi}{6}i}} \frac {z^4}{z^6+1} (z-e^{\frac {\pi}{6}i})=\frac {e^{\frac {4\pi}{6} i}}{6e^{\frac {5\pi}{6}i}} = \frac {e^{-\frac {\pi}{6}i}}{6}$

Now that we have one residue, the rest get easier. To find the residue at $e^{\frac {3\pi}{6}i}$ we can quickly jump through the algebra above to

$\lim_\limits{z\to e^{\frac {3\pi}{6}i}} \frac {z^4}{z^6+1} (z-e^{\frac {3\pi}{6}i}) = \frac {e^{\frac {12\pi}{6}i}}{6e^{\frac {15\pi}{6} i}} = \frac {e^{-\frac {3\pi}{6}i}}{6}$

And the pattern becomes more obvious. The rest of the residues are:

$\frac {e^{-\frac {5\pi}{6}i}}{6}, \frac {e^{-\frac {7\pi}{6}i}}{6},\frac {e^{-\frac {9\pi}{6}i}}{6}, \frac {e^{-\frac {11\pi}{6}i}}{6}$