# Principal part calculation

I just wanted to find the principal part of

$$f(z) = \frac{1}{(1-z^3)^2}$$

by calculating the negative coefficients of its Laurent series expansion applying

$$a_{n} = \frac{1}{2\pi i}\oint \frac{f(z)dz}{(z-z_0)^{n+1}}$$

with the singularity at $z_0 = 1$, but I got stuck quite at the beginning with a fourth order polynomial in z squared in the denominator etc. So I thought what I'm doing is probably wrong ...

Did I try the right approach after all, or is there a trick / much simpler / better way how to find the principle part in this case ?

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Geometric series and known MacClaurin series are much faster than explicit computation of integrals in many cases. The answer below is worth thinking about for other problems you work as well. Think about second semester calculus, similar comments apply: do you want to calculate the Talyor series for $x^3e^{x^2}$ by direct differentiation or by subsitution of $x^2$ into the known expansion for $e^x$? Many interesting questions of complex variables involve integrating without integrating. This is how. – James S. Cook Oct 6 '12 at 20:58

The principal part is for when $|z| >1$. Rewrite as $\frac{1}{z^6(1-\frac{1}{z^3})^2}$. Then write out the geometric series for $\frac{1}{1-\frac{1}{z^3}}$ and square it.