Find $\int_{|z|=1} f(z) d z$, when $f(z)=(z \sin z) /(z+2)+\bar{z}$
$f(z)$ looks like analytic function. If it is analytic, then by Cauchy-Theorem, integration will be zero. I tried to show Cauchy-Riemann equations by using $$\frac{\partial f}{\partial \bar{z}}=0,$$
I got $$\frac{\partial }{\partial \bar{z}} ((z \sin z) /(z+2)+\bar{z}) = 1 + \left(\frac{\partial }{\partial \bar{z}} (z \sin z) /(z+2)\right)$$
I don't know how to proceed.
Since $|z| = 1$, I replaced $z = 1 / \bar{z}$ and $z/(z+2) = 1 / (2\bar{z} + 1)$, but I am not sure if it is the right way.
Best Answer
You can try this,
$$\oint_{|z|=1}\left(\frac{z\sin z}{z+2}+\overline z\right)dz=\oint_{|z|=1}\overline z\,dz$$
and now write (by mere definition of complex line integral, with parametrization and etc.)
$$\;z=e^{it}\;,\;\;0\le t\le 2\pi\implies dz=ie^{it}dt$$
so the remaining integral equals 0 $$\int_0^{2\pi}e^{-it}\,i\,e^{it}\,dt=i\int_0^{2\pi}dt=2\pi i$$
Observe that the first part of the first integral (the first summand's integral) is zero since that is an analytic function on the unit closed disk.