Is f(z)=1/z truly an analytic function For an analytic function $f(z)$, we have $$\frac{\partial f}{\partial \bar{z}}=0.$$ Consider the function $f(z)=\frac{1}{z}$, which, at first sight, is a bona fide analytic function. However, we can write it as $$f(z) = \frac{\partial}{\partial z} (\ln z + \text{const}),$$ where const is any expression that doesn't depend on $z$. So, let's take $\text{const}=\ln \bar{z}$ so that $$f(z) = \frac{\partial \ln z\bar{z}}{\partial z}.$$ Now the condition that $f(z)$ is analytic takes the form $$\frac{\partial^2}{\partial z \partial \bar{z}} \ln z \bar{z} = 0.$$ However, by employing the cartesian coordinates, we see that this is exactly the 2D Green function for laplacian $$\frac{1}{4} (\partial_x^2 + \partial_y^2) \ln (x^2 + y^2) = \pi \delta(\vec{r}).$$
My question: how can a perfectly decent analytic function $f(z)=\frac{1}{z}$ secretly depend on $\bar{z}$? I presume that there is a deep connection with the residue theorem?
 A: A word of caution: $\dfrac{1}{z} \neq \dfrac{\partial}{\partial z}\log(z)$. These are inherently different functions due to domain of definition. $\log(z)$ is only well-defined if you make a branch cut, which means that you are eliminating an entire ray emanating from the origin from your domain but $\dfrac{1}{z}$ is well-defined for all $z\neq 0$. This can likely be remedied by an Identity Theorem type argument though so it's not a big issue.
Furthermore, $\frac{1}{z}$ is not holomorphic everywhere and so we can expect weird things to happen sometimes which is the crux of your post. When introducing $\log(\bar{z})$, you are implicitly assuming that $\bar{z}\neq 0$. In this setting, what you get is that the Laplacian of $\log(z\bar{z})$ is zero which you should expect. If $z=0$, none of what you did makes sense from the get-go (since $\log(\bar{z})$ is undefined!) so your oddities really only exist because you're not treating your functions as you should.
To add on to this post: the appearance of the Dirac delta is not surprising and does indeed get reflected in the Cauchy integral formula. You can view the Cauchy integral formula as the line integral portion of Green's (first? second?) theorem (from which the Green's function appears quite naturally). Saff does this treatment which really gives complex analysis a huge boost because it relates directly to vector analysis in some ways.
Really you should treat the Dirac delta not as a function but as what it does in integration. As a function, $\dfrac{\partial^2}{\partial z\partial\bar{z}}\log(z\bar{z}) = 0$ but when you do closed contour integrals of functions against $\log(z\bar{z})$ you end up not getting zero which is where the Dirac delta property comes into play and the Dirac delta property all in one fell swoop.
A: If f is function like $Re^{i \theta}$, that constant can vary depending on your choice for domains for the angle, and thus not be equal to $\bar z$. They can map similarly, but the functions are not equal.
