Laurent Series of $~\tan(z)~$ expanded in $\frac{\pi}{2} < |z| < \frac{3\pi}{2}~$? As we know, we can get Laurent series of $~\tan(z)~$ expanded in $~0 \le |z| \lt \frac{\pi}{2}~$ by dividing Taylor series expansion  of $~\sin(z)~$ by Taylor series expansion of $~\cos(z)~$, and we'll get $$\tan(z) = \frac{\sin(z)}{\cos(z)} = \frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}+\dots}{1-\frac{z^2}{2!}+\frac{z^4}{4!}+\dots} = z+\frac{z^3}{3}+\frac{2z^5}{15}+\dots$$
Now I am curious about the Laurent series of $~\tan(z)~$ expanded in $~\frac{\pi}{2}~< |z| < \frac{3\pi}{2}$.
Since Taylor series of $~\sin(z)~$ and $~\cos(z)~$ is valid everywhere on the complex plane, I originally think the answer will be the same as above.
But one exercise on Brown&Churchill's Complex variables and applications states that value of the integral 
$$\oint\limits_C \tan(z)\,dz = -4\pi i$$
where path $~C~$ is the positively oriented circle  $~|z| = 2~$,  which lies in the domain $~\frac{\pi}{2}~ < |z| < \frac{3\pi}{2}~$. 
If Laurent series of $~\tan(z)~$ expanded in $~\frac{\pi}{2}~ < |z| < \frac{3\pi}{2}~$ is really $$z+\frac{z^3}{3}+\frac{2z^5}{15}+\dots$$, then the residue (coefficient of order $~-1~$) is zero, which means the integral should be $~0~$, contradicting the result of the exercise. 
So it is clear that the above series form is not the correct Laurent series of $~\tan(z)~$ in domain $~\frac{\pi}{2}~ < |z| < \frac{3\pi}{2}~$.
Then what is the correct Laurent Series of $~\tan(z)~$ in domain $~\frac{\pi}{2}~ < |z| < \frac{3\pi}{2}~$? 
And can somebody help me explain why I can't use Taylor Series of $~\sin(z)~$ and $~\cos(z)~$ to derive the Laurent Series? 
What's wrong with my concept?
Thank you very much!
 A: The Laurent series of $f(z)$ in an annulus $a < |z| < b$ (assuming the function is analytic there) is $\sum_{n=-\infty}^\infty c_n z^n$ where
$$ c_n = \dfrac{1}{2\pi i} \oint_C f(z) z^{-n-1}\; dz $$
for a simple closed positively-oriented contour $C$ that goes around $0$ in this annulus.
In your case, you can compute this using residues: the poles to consider are $0$ and $\pm \pi/2$.  The residues at $0$ give you the same coefficients as you would get for the Maclaurin series.  The residues at $\pm \pi/2$ give you something new.
EDIT: Another way of looking at it: $g(z) = \tan(z) + \frac{1}{z-\pi/2} + \frac{1}{z+\pi/2}$ is analytic in $|z|<3\pi/2$ (check that the singularities at $\pm \pi/2$ are removable).  Its Maclaurin series is the sum of the Maclaurin series of $\tan(z)$, $1/(z-\pi/2)$ and $1/(z+\pi/2)$.  To get a series for $f(z)$, you need to subtract the $1/(z-\pi/2)$ and $1/(z+\pi/2)$, and to do this in a way that is
convergent in your annulus, you must use their Laurent series for $|z| > \pi/2$, which involve negative powers of $z$.
