Complex integral - exercise $$ \int\limits_{C(-2, \frac{1}{4})} = \frac{e^z}{z^2-4}dz$$
C is a circle center = -2 and radius = $\frac{1}{4}$ z is a complex number
I don't know how to do the exercises like that. 
 A: It is important to specify the direction of the contour.
$$
\int_\gamma\frac{e^z}{z^2-4}\,\mathrm{d}z
=\underbrace{\frac14\int_\gamma\frac{e^z}{z-2}\,\mathrm{d}z}_{I_1}
-\underbrace{\frac14\int_\gamma\frac{e^z}{z+2}\,\mathrm{d}z}_{I_2}
$$
where $\gamma$ is a counterclockwise circle with center $2$ and radius $\frac14$. 
$\gamma$ circles the simple pole of the integrand of $I_1$ at $z=2$ once, but does not contain the pole of the integrand of $I_2$ at $z=-2$.
We can evaluate $I_1$ using Cauchy's Integral Formula and $I_2$ using Cauchy's Integral Theorem.
A: Let D be a simply connected domain bounded by a contour C. By Cauchy's integral formula, if f is analytic in D and 
$z_0 \in D$ then
$$\frac{1}{2\pi i}\int_C^\ \frac{f(z)}{z-z_0}dz\ = f(z_0). $$
Note also that  for $z_0 \not\in D,$ 
$g(z) =\frac{f(z)}{z-z_0} $ is analytic in D for $ z_0 \not\in D$. In particular, by Cauchy's theorem, for $z_0\not\in D,$
$$ \int_C^\ \frac{f(z)}{z-z_0}dz\ =0. $$
Let us take a couple of examples. Let C be the circle, $|z| = 3$. Consider the integrals
$$ I_1 = \int_C^\ \frac{e^z}{z-1}dz $$ 
$$ I_2 = \int_C^\ \frac{e^z}{z-4}dz\ $$
We note that $e^z$ is analytic so we can apply Cauchy's integral formula. Since the point $z=1$ is contained in C, we have
$$\frac{1}{2\pi i}\int_C^\ \frac{e^z}{z-1}dz\ = e^1. $$
so $I_1 = 2\pi i e$. In the case of $I_2$, we see that the singularity $z=4$ is not contained with the contour C. Hence $\frac{e^z}{z-1}$ is analytic in the region bounded by C, so, by Cauchy's theorem,
$$ I_2 = \int_C^\ \frac{e^z}{z-4}dz\ =0 $$
In your case we note that we can write 
$$\frac{e^z}{z^2-4} = \frac{\frac{e^z}{z-2}}{z+2} = \frac{f(z)}{z+2} $$
where $f(z)=\frac{e^z}{z-2}$ is analytic in your C. Since $z=-2$ is contained within C, we apply Cauchy's integral formula,
$$\frac{1}{2\pi i}\int_C^\ \frac{f(z)}{z+2}dz\ = f(-2). $$
$$\int_C^\ \frac{f(z)}{z+2}dz\ = 2\pi i f(-2)=-2\pi i\frac{e^{-2}}{4}. $$
A: Consider
$\frac{e^z}{z^2-4} = \frac{1}{z+2} \frac{e^z}{z-2}$. Now call $f(z)=\frac{e^z}{z-2}$. Then the given integral can be rewritten as
$\int_C \frac{f(z)}{z+2} \, dz$. Observe that $f(z)$ is analytic on the given domain. Use the residue theorem.
A: Zeros: $z=-2,2$
Residue Theorem: $\int_{C}\frac{f(z)}{z-z_0} dz = 2 \pi i \sum Res(f,z_0)$
Note: We can just compute: $\int_{C} \frac{\frac{e^z}{z-2}}{z-(-2)} dz = 2 \pi i (f(z)|_{z=-2})$, where $f(z)=\frac{e^z}{z-2}$.
A: You must use residue theorem: it says that the integral along a circular path is equal to $$ 2\pi i \sum Res(f,z_0)$$ where $z_0$ is a point of singularity inside your path.
In your case the points of singularity are the solution of $$z^2-4=0 \quad\rightarrow\quad  z=\pm 2$$
Pay attention: your path is a circle center in $-2$ with radius $\frac14$, so you have only one singularity point: $z=-2$.
Now you have to evaluate the residue in $z=-2$; suppose that $z_0$ is a singularity point and that you can write your function as $$g(z) =\frac{f(z)}{z-z_0}$$ then the residues in $z_0$ is equal to $f(z_0)$. In your case:$$ \frac{e^z}{z^2-4}=\frac{e^z}{(z-2)(z+2)} = \frac{\frac{e^z}{z-2}}{z+2} = \frac{\frac{e^z}{z-2}}{z-(-2)} $$ $$\Longrightarrow\quad Res(f,-2)= \frac{e^-2}{-4}$$
Now you can simply calculate your integral:
$$ \int\frac{e^z}{z^2-4} = -2\pi i\frac{e^-2}{4} $$
