Calculating a complex integral $\int_{\gamma}\frac{4\cos z}{z}dz$

Solution 1:

You should look at the Residue theorem. It says that in a simply connected region bounded by a contour($\gamma$) such as this where the function $f$ is analytic in the region except for a finite number of points $\{a_{1},a_{2},...,a_{n}\}$.

$$\int_{\gamma}f(z)\,dz=2i\pi\sum_{i=1}^{n}\text{Res}_{z=a_{i}}f(z)$$.

Using this you get your answer correctly $2i\pi\text{Res}_{z=0}f(z)=2i\pi\cdot 4 = 8i\pi$.

Essentially if you want to make more sense out of it.

Then you consider the Laurent series expansion of $f(z)=\frac{4\cos(z)}{z}$. about $z=0$.

You get $\displaystyle 4\left(\frac{1}{z}-\frac{z}{2!}+\frac{z^{3}}{4!}-...\right)$.

And you use the fact that when you integrate on the contour $\gamma(|z|=7)$. You get the integrals of the form $\int_{\gamma}z^{n}\,dz=0$ for $n\in\mathbb{Z}\setminus\{-1\}$. and the integral is $2i\pi$ when $n=-1$. Using this you can sort of make sense of what's happening.

Solution 2:

The function $f(z)=\frac{4 \cos z}{z}$ has a pole of order 1 at $z = 0$. Using the residues theorem, you can compute the integral $\int_{\gamma} f(z)dz$, whenever $0 \notin \gamma$. For simplicity, if $\gamma$ is a simple curve we have three possibilities:

1. $0$ is in the interior of $\gamma$ and the value of the integral is given by the residues theorem (it does not depend on $\gamma$).

2. $0$ is in the exterior of $\gamma$ and the value of the integral is zero (again, the integral does not depend on $\gamma$).

3. $0$ is on $\gamma$ and you need to study the situation more carefully. This is surely the only way you can obtain $11$ as the value of the integral.

Edit: It was not stated in the OP that $\gamma$ needed to be a closed contour... And my answer was based on this assunption. If it can be for instance a line segment, finding a path for which the integral is $11$ is quite trivial, and points 1. and 2. do not apply.