# Determine the type of singularity and principal part of any pole at z=0

We have $$f(z) = \frac{1}{z\ln(\cos z)}$$ I determined the singularity to be a pole of degree 3 by taking $$\lim_{z\to 0}f(z)(z-0)^3=-2$$ However how do I go about finding the principal part? Is there a trick to expand the function using what I know? I've tried starting by expanding $\ln(z)$ and plugging in $\cos(z)$ and so forth just to end up with a jumbled mess and getting nowhere. Any help is appreciated. I'm really not sure how to go about this.

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"finding the principal branch"...of whom, of what or what? Functions have branches (sometimes), not points. Perhaps you meant the p. branch of the logarithm function, which you obtain by "erasing" the non-positive real axis (and thus positive real numbers have argument = 0)...? – DonAntonio Mar 18 '13 at 12:43
Sorry about that> I typed it correctly in the title, then i guess my mind was wandering. I meant how can I go about finding the principal part, not branch. – Pauli Mar 18 '13 at 19:40
Maybe you are asking for the principal part of the Laurent series. – Mhenni Benghorbal Mar 18 '13 at 19:45
So apparently you mean "the principal part of the function" as with Laurent series and stuff, uh? – DonAntonio Mar 18 '13 at 19:46

We need to use the Taylor expansions $\cos z=1-z^2/2+z^4/24+O(z^6)$, $\ln (1+u)=u-u^2/2+O(u^3)$, and $(1+v)^{-1}=1-v+O(v^2)$ at $0$.
It follows that $$\ln\cos z=-\frac{z^2}{2}+\frac{z^4}{24}- \frac{z^4}{8}+O(z^6)=-\frac{z^2}{2}- \frac{z^4}{12}+O(z^6)=-\frac{z^2}{2}(1+\frac{z^2}{6}+O(z^4))$$ $$z\ln\cos z=-\frac{z^3}{2}(1+\frac{z^2}{6}+O(z^4))$$ and $$f(z)=\frac{1}{z\ln\cos z}=-\frac{2}{z^3}(1+\frac{z^2}{6}+O(z^4))^{-1}=-\frac{2}{z^3}(1-\frac{z^2}{6}+O(z^4)).$$ So
$$f(z)=-\frac{2}{z^3}+\frac{1}{3z}+O(z).$$ You have your principal part. Check: $f$ is odd, and so is what we found. So this can't be awfully wrong. Double check: this is correct.
The principal branch, $\text{Log}{z}$, is defined such that $\text{Log}{1}=0$, which you assumed in your residue calculation.