Expand $\cot z$ around $z_{0} = n\pi$

Question

I'm trying to expand the function in a Laurent series

$$f(z) = \cot z$$

around the singularity $$z_{0} = n\pi$$ Initially I tried expanding $\tan z$ with a Taylor expansion:

$$\tan(z) = \tan(n\pi) + \tan'(n\pi) (z - n\pi) + \dfrac{tg''(n\pi)}{2}(z - n\pi)^{2} + \dfrac{tg'''(n\pi)}{6}(z - n\pi)^{3} + ...$$

$$\tan(z) = \dfrac{(z - n\pi)}{\cos(n\pi)} + \dfrac{1}{3}\dfrac{(z - n\pi)^{3}}{\cos^{4}(n\pi)} + ...$$

To obtain the inverse:

$$\cot z = \dfrac{1}{\tan z} = \left[\dfrac{(z - n\pi)}{\cos(n\pi)}\left(1 + \dfrac{1}{3}\dfrac{(z - n\pi)}{\cos^{3}(n\pi)}\right)\right]^{-1}$$

But I can't work from here

Answer 1

Hint

You were very correct o start with the expansion of the tangent around $n\pi$.

Truncated to some order $$\tan(z)=(z-\pi n)+\frac{1}{3} (z-\pi n)^3+\frac{2}{15} (z-\pi n)^5+\frac{17}{315} (z-\pi n )^7+O\left((z-\pi n)^{9}\right)$$

Now, using $$\cot(z)=\frac 1 {\tan(z)}$$ just perform the long division