Calculate:$\int _{|z|=1}\text{Re z dz}$. For a positively oriented unit  circle 

Calculate:$\int _{|z|=1}\text{Re z dz}$.

If $z=x+iy\implies \text{Re z}=x$
So $\int _{|z|=1}\text{Re z dz}=\int_{|z|=1} x d(x+iy)=\int _{|z|=1}x dx$.(as dy=0)
$=\frac{x^2}{2}|_0^{2\pi }=2\pi^2$
But the answer is not matching.It is given as $\pi i$.
Where am I wrong ?Please help.

On the unit circle $\gamma$ one has $\bar z={1\over z}$. It follows that
$$\int_\gamma {\rm Re}\,z\>dz={1\over2}\int_\gamma(z+\bar z)\>dz={1\over2}\int_\gamma z\>dz+{1\over2}\int_\gamma{1\over z}\>dz=0+{1\over2}2\pi i=\pi i\ .$$
