Evaluating a complex integral over a half-ring I need to integrate the $z/\bar z$ (where $\bar z$ is the conjugate of $z$) counterclockwise in the upper half ($y>0$) of a donut-shaped ring. The outer circle is $|z|=4$ and the inner circle is $|z|=2$. 
My method:
$z/\bar z = e^{2i\theta}$ - which is entire over the complex plane.
So with respect to $d\theta$, we get the integral $re^{i3\theta} d\theta$ which, we can then evaluate at r=4 (from pi to 0) and r=2 from (0 to pi)
Two questions:
1) As integrating in the counterclockwise direction, surely I shouldn't be getting a negative number?
2) Via the deformation theorem, as the function is holomorphic on both circles and the region between them, should I not be getting 0? 
 A: In the ccw direction, there are 4 contributions to this integral:
$$\begin{align}\oint_C dz \frac{z}{z^*} &= 4 \int_0^{\pi} d\theta \: e^{i 3 \theta} -  2\int_0^{\pi} d\theta \: e^{i 3 \theta} + \int_{-4}^{-2} dt + \int_2^4 dt\\ &= \frac{-8}{3 i} + \frac{4}{3 i} + 2 + 2 \\ &= 4 + i\frac{4}{3} \end{align} $$
The fact that this is $\ne 0$ has something to do with the fact that $z^*$ is not holomorphic within the integration region.
A: 1) $e^{2i\theta}$ is not holomorphic, and therefore not entire. There are many way to check this, but it suffices to observe that $\frac{\partial}{\partial \bar{z}}\frac{z}{\bar{z}} = -\frac{z}{\bar{z}^2}\neq 0$. See the discussion of the Wirtinger derivative in the definition section here: wikipedia.
2)The deformation theorem you refer to is about integrating holomorphic functions over contours. It looks like you are trying to evaluate the area integral:
\begin{equation}
\int_{r=2}^{r=4}\int^{\theta=\pi}_{\theta=0}z/z^{*}rdrd\theta
\end{equation}
So even if the function were holomorphic, you would not neccessarily get zero.
A: @rlgordonma Thank you for your help! Just a quick question: I do the same integration as you but I seem to end up integrating rie^3itheta rather than just re^3itheta (as dz=d(re^i3theta)=rie^i3theta dtheta)
Which is right?
