$\int \frac{2+\sin(z)}{z} dz$

Question

Please bear in mind that I am trying to teach myself complex integration having never taken a course in complex analysis, so assume I know very little.

$\int \frac{2+\sin(z)}{z} dz$ traversing the unit circle once counterclockwise.

I looked up cauchy's integral formula, but I don't know how to apply it. What do I do?

I'm getting $4 \pi i$ by saying $\int \frac{2+\sin(z)}{z} dz = 2+ \sin(0) $but i'm pretty sure that's wrong.

Answer 1

Let the contour $\gamma$ be $$\gamma(t) = e^{it},\; 0 \leq t \leq 2\pi $$ We want to calculate the result of the contour integral

$$\oint_{\gamma}\frac{2 + \sin(z)}{z} \; \text{d}z$$

With the theory of complex residues, a contour integral $\oint_{\gamma} f(z) \; \text{d}z$ is equal to $2\pi i$ times the sum of all complex residues of the singularities $z_i$ of the function $f(z)$ which are inside the closed curve $\gamma$. That is

$$\oint_{\gamma} f(z) \; \text{d}z = 2\pi i \sum_{z_i} \text{Res}(f, z=z_i)$$ Looking at your integrand $$f(z) = \frac{2+\sin z}{z}$$ We see that there is a simple pole (of degree $1$) at $z = 0$. Since $\gamma$ is the closed curve describing the unit circle, the point $z = 0$ lies within that unit circle. So we have

$$\oint_{\gamma}\frac{2 + \sin(z)}{z} \; \text{d}z = 2\pi i\cdot \text{Res}(f, z_0=0)$$

For this, we must now calculate the complex residue of the singularity of $f$ at $z_0=0$. Using the formula on wikipedia, we have for a simple pole at $z_0$:

$$\text{Res}(f, z_0 =0) = \lim_{z \to z_0} (z-z_0)f(z)$$

Substituting that in, we get

$$\lim_{z \to z_0} (z-z_0)f(z) = \lim_{z \to 0} \; 2+\sin z = 2 + \sin(0) = 2 $$

And as such the residue of the singularity of the function $f(z)$ at $z=0$ is equal to $2$. We get the result of the contour integral as

$$\oint_{\gamma}\frac{2 + \sin(z)}{z} \; \text{d}z = 2\pi i\cdot 2 = 4\pi i$$

Answer 2

1/ You have to search the singularities in the contour.

if the singularities are pole, you have to search the residue which is the coefficient of 1/(z-a) where a is the pole;

2/ beware to the branch point like log(z) in z=0 and sqrt(z) in z=0 , and others. In these cases there is no residue and this is more complicated.

3/ if you have only poles, you apply the residue theorem: the pole a of residue R have a contribution equal to $2i\pi R$ (I expect that you make only one circle around the pole a).

Here: you have $f(z) = \frac{2+sin(z)}{z}$. the function have a singularity in z=0 as the numerator is 2 and the denominator is 0 in $z=0$. so the function is like $f(z) = \frac{2}{z}+1+ O(z^2)$ so the residue is 2.

the residue theorem give the value $4i\pi$.

Answer 3

CIT tells us that if $\;f(z)\;$ is analytic on a closed, simple and rectifiable path $\;\gamma\;$ and $\;a\;$ is an interior point to the domain enclosed by this path, then

$$f(a)=\frac1{2\pi i}\oint_\gamma\frac{f(z)}{z-a}dz\implies\oint_\gamma\frac{f(z)}{z-a}dz=2\pi if(a)$$

In your case, with $\;\gamma=S^1:=\{z\in\Bbb C\;/\; |z|=1\}\;$ and $\;f(z)=2+\sin z\;$, check all the conditions for CIT are fulfilled and thus

$$\oint_{\gamma}\frac{2+\sin z}zdz=2\pi i(2+\sin z)|_{z=0}=2\pi i(2+0)=4\pi i$$