Poles and Laurent series of $\tan z$ I am very confused on how to obtain the principal part, and in general the Laurent series of functions of complex variable.
I will write an exercise and try to point out where are my doubts.

Consider the function $\tan(z)$ in the annulus $\lbrace3<|z|<4\rbrace$. Let $f(z)=f_0(z)+f_1(z)$ be the Laurent decomposition of $f(z)$, so that $f_0(z)$ is analytic for $|z|<4$ and $f_1(z)$ is analytic for $|z|>3$ and vanishes at $\infty$. (a) Obtain an explicit expression for $f_1$.

Since $\cos(z)=0$ only when $z=\pm(2n+1)\pi/2$ ($n=0,1,2,\dots$), there are no poles inside the annulus. Therefore $f_1(z)=0$. Is this correct?

(b) Write down the series expansion for $f_1(z)$ and determine the largest domain on which it converges.

If I am wrong before, what is the answer to this?
 A: Hints: No, $f_1$ is not zero. If it were, then $\tan(z) = f_0(z)$ would be analytic for $|z| < 4.$
$f_1(z)$ needs to account for the poles within $|z| \le 3.$ There are two: one at $\frac{\pi}{2}$ and the other at $-\frac{\pi}{2},$ and both are simple. So your expression should be something like $$f_1(z) = \frac{C_1}{z - \pi/2} + \frac{C_2}{z + \pi/2}$$ for some constants $C_1,C_2.$
A: It is true that there are no poles inside the annulus $3 < \lvert z \rvert < 4$. But there are poles in the disk $\lvert z \rvert <3 $ (at $z=\pm \pi/2$), so we have to include these in $f_1$ so that $f_0$ can be analytic there. $f_1$ has no poles outside $\lvert z \rvert <3$, so we need to find out how to cancel out the poles of $\tan{z}$ at $\pm \pi/2$.
We know that 
$$\tan{z} = \cot{(\pi/2-z)} = \frac{1}{\tan{(\pi/2-z)}}.$$
Also, $\tan{w}/w \to 1$ as $w \to 0$, so 
$$ \frac{\pi/2-z}{\tan{(\pi/2-z)}} \to 1 \quad \text{as } z \to \pi/2. $$
Hence $z=\pi/2$ is a simple pole, so $\tan{z} - \frac{1}{\pi/2-z} $ has a removable singularity at $z=\pi/2$. Since $\tan{z}$ is odd, we know that there is another pole at $z=-\pi/2$, and a similar calculation shows that $\tan{z}-\frac{-1}{z+\pi/2}$ is has a removable singularity at $z=-\pi/2$. Therefore,
$$ \tan{z} - \frac{1}{\pi/2-z} - \frac{-1}{\pi/2+z} = \tan{z} - \frac{8z}{\pi^2-4z^2} $$
is analytic for $\lvert z \rvert < 3$. Therefore this is $f_0$, and
$$ f_1(z) = \frac{8z}{\pi^2-4z^2} $$
For the second part, presumably it means the expansion for $\lvert z \rvert > 3$:
$$ \frac{8z}{\pi^2-4z^2} = -2z^{-1} (1-\frac{\pi^2}{4z^2})^{-1}, $$
which I'm sure you can expand using the geometric series formula. In particular, this converges for $\lvert \pi^2/(4z^2) \rvert < 1$.
