I have this integral to calculate:
$I=\int_{|z|=2}(e^{\sin z}+\bar z)dz.$
I do it this way:
$I=\int_{|z|=2}e^{\sin z}dz+\int_{|z|=2}\bar zdz.$
The first integral is $0$ because the function is holomorphic everywhere and it is a contour integral. As for the second one, I have
$\int_{|z|=2}\bar zdz = \int_0^{2\pi}e^{-i\theta}\cdot 2 d\theta=-\int_0^{-2\pi}e^{i\tau}\cdot 2 d\tau=\int_0^{2\pi}e^{i\tau}\cdot 2 d\tau=\int_{|z|=2}zdz=0$
because the function is now holomorphic.
It seems fishy to me. Is it correct?