<s>[INST] Residue Computation How can we compute the residue of the function $\dfrac{(1+zx)^{k+1/3}}{z^2(z-a)} $ at $z=0$?? where $x$ is some fixed real number and $k$ is some fixed positive integer? I have tried computing laurent series of each but it gets too complicated to do.
 [/INST] The residue of $\frac{f(z)}{z^2}$ at $z = 0$, where $f$ is holomorphic near $0$ is $f'(0)$, or, which sometimes is easier to find, the coefficient of $z$ in the Taylor expansion of $f$. Here we have
$$f(z) = \frac{(1+zx)^c}{(z-a)},$$
which is not totally trivial to differentiate (but not too hard either). We can develop the numerator into a binomial series
$$(1+xz)^c = \sum_{\nu=0}^\infty \binom{c}{\nu}(xz)^\nu = 1 + c\cdot xz + \frac{c(c-1)}{2}x^2z^2 + \dotsb$$
and the other factor into a geometric series
$$\frac{1}{z-a} = \frac{(-1)}{a} \frac{1}{1- \frac{z}{a}} = \frac{(-1)}{a} \sum_{\kappa=0}^\infty \left(\frac{z}{a}\right)^\kappa = -\frac{1}{a} - \frac{z}{a^2} - \frac{z^2}{a^3} - \dotsb$$
Now we need the coefficient of $z^1$ in the product of the two series,
$$-\left(\frac{1}{a} + \frac{z}{a^2} + \dotsb\right)\left(1 + cxz + \dotsb\right) = -\frac1a - \left(\frac{1}{a^2} + \frac{cx}{a}\right)z + \dotsb$$
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