Calculating the Laurent Series of $\tan z$

I need help calculating the laurent series of $\tan z$ around the points $z=0$, $z=\pi/2$, and $z=\pi$.

How would one go about doing this? I solved an almost identical question that was "Derive the Laurent expansion for the function $f(z)= \frac{\sin z}{\cos z}$ around $z=\pi/2$. Use long division."

Is it the same here?

• You can literally do this problem like this post of yours – dustin Nov 16 '14 at 22:09
• @dustin so the power series is the same as the Laurent series in this case? – User38 Nov 16 '14 at 22:21
• The Taylor series is for $n\geq 0$ and a Laurent series has $n < 0$. Tangent already has a well known Taylor series for $n\geq 0$. So what you are doing is expanding the Taylor series around some point like this Calculus nothing to do with Complex variables. – dustin Nov 16 '14 at 22:28