I'm confused about how to calculate $\nabla^2 \log z$, where $z=re^{i\theta}$ is a complex number. My calculations return $$ \nabla^2 \log z = 2\pi\frac{\delta(r)}{r} [\delta(\theta) + i \Theta(\theta)] $$ , where $\delta$ is the dirac delta and $\Theta$ is the step function (Heaviside).
I can find no errors in the above expression, but when I try to check the above result by using vector calculus, I have on the left side $$ \int_A da \nabla^2 \log z = \mbox{using stokes theorem} = \int_{\partial A} d\vec{l} \cdot \vec{\nabla} \log z = 2\pi $$ where I choose $A$ a circle of radius $A$. By the other hand, the right hand side gives $$ 2\pi(1+2\pi i) $$ Does anyone has a clue about what am I doing wrong?
EDIT
As I was discussing with geodude below, my main concern is the correctness of the imaginary part of $\nabla^2 \log z$. The real part is ok, because it appears everywhere in electrostatics and I've verified that.
So, to obtain the full solution, I've performed as following:
Given $\log(z-z')$, since it is singular at $z=z'$, I've used a series expansions for the log. $\log(z-z') = (A-B) \Theta(r-r') + B$, where $$ A = \log r + i \theta - \sum_{m=1}^\infty \frac{1}{m} \left(\frac{r'}{r}\right)^m e^{im(\theta'-\theta)} \\ B = \log r' + i \theta' - \sum_{m=1}^\infty \frac{1}{m} \left(\frac{r'}{r}\right)^m e^{im(\theta-\theta')} $$
Since $\nabla^2 B = 0$, only the term involving (A-B) must be calculated, and after turning the machinery I've obtained the equation shown at the top.