Integral using Cauchy's integral formula

Question

How can we compute the integral of $ \int_{|z|=1}f(z)dz$ , where $f(z) = z \sin(z) / (z+2) + \bar{z} $

I think we need to use Cauchy's integral formula, but I am not sure how?

Answer 1

Let $\gamma$ denote the unit circle parametrised in the anti-clockwise direction. Split: $$ \int_\gamma f(z) dz = \int_\gamma \frac{z \sin z}{z + 2} dz + \int_\gamma \overline{z} dz $$ The former is holomorphic within $\gamma$, so...

The second integral should be easy to compute.

Answer 2

Let $D:=\{z \in \mathbb C:|z|<3/2\}$. Then $g(z):=z*sin(z)/(z+2)$ is holomorphic on $D$.

Thus (Cauchy !): $\int_{|z|=1}g(z)dz=0$

A direct calculation gives $\int_{|z|=1} \overline{z}dz=2 \pi i$