Residues theorem, and integrate 
Evaluate $\int_\gamma\frac{z}{z^2+2z+5}dz$ where $\gamma$ is the unit
  circle

I did but I don't know if it's right
$$z^2+2z+5=0\Leftrightarrow z=-1\pm 2i$$
this I have that $z_1=-1+2i$ and $z_2=-1-2i$ are the singularity points, then 
$$\int_\gamma \frac{z}{z^2+2z+5}dz=2\pi i*\sum_{i=1}^n [Res(f;z_i)]I(\gamma;z_i)$$
where $I(\gamma;z_i)=1$ if $z_i$ it is inside of $\gamma$ and $I(\gamma;z_i)=0$ if $z_i$ it is outside of $\gamma$. In that case $z_1$ and $z_2$ are outside of $\gamma$, so $$\int_\gamma \frac{z}{z^2+2z+5}dz=2\pi i*[0+0]=0$$
My reasoning is correct?

EDIT: Evaluate $\int_\gamma \tan z \space dz$ where $\gamma$ is the
  circle of radius $8$ centered at $0$

Someone could help me with this, I'm not thinking through solve it without using series.

HINT 1:
The poles of $\tan z$ inside the circle $|z|=8$ are located at $\pm \pi/2$, $\pm 3\pi/2$, and $\pm 5\pi/2$.  The poles are of order $1$ and the residue at $(2n+1)\pi/2$ is
$$\lim_{z\to (2n+1)\pi/2}(z-(2n+1)\pi/2)\tan z$$
HINT 2:
To find the limit $\lim_{z\to (2n+1)\pi/2}(z-(2n+1)\pi/2)\tan z$ we use L'Hospital"s Rule.  To that end
$$\begin{align}
\lim_{z\to (2n+1)\pi/2}(z-(2n+1)\pi/2)\tan z&=\lim_{z\to (2n+1)\pi/2}\frac{(z-(2n+1)\pi/2)\sin z}{\cos z}\\\\
&=\lim_{z\to (2n+1)\pi/2}\frac{\sin z+(z-(2n+1)\pi/2)\cos z}{-\sin z}\\\\
&=-1
\end{align}$$
