Showing a region where a functions is harmonic Let $f(x,y) = \frac{ y}{(x-1)^2+y^2} $. I want to find the region where $f$ is harmonic.
Try:
We know $f$ is analytic everywhere except at $(1,0)$. Hence, $f$ is harmonic everywhere except on $(1,0)$. Is this correct? or do I have to carry out the derivations?
 A: Let $\overline{h(z)} = \frac{1}{z-1}$, then $h$ is anti-holomorphic for $z \neq 1$, so $h$ is harmonic if $z \neq 1$. 
Also $\Im(h)(x+iy) = \frac{y}{(x-1)^2 +y^2} $...
Or you could brute-force it in a similar manner to How to find the set for which a function is harmonic
A: One does not need to establish that $f=\frac{y}{(x-1)^2+y^2}$ is harmonic by taking partial derivatives directly.  If we observe that the function $f$ given by 
$$\text{Im}\left(\frac1{1-z}\right)=\frac{y}{(x-1)^2+y^2}$$
and if we know that $f(z)$ is analytic everywhere in the punctured plane (i.e., the plane less the point $(1,0)$), then both its real and imaginary parts are harmonic functions.  So, we are done.
We can prove the general result that the real and imaginary parts of an analytic function $f(z)=u(x,y)+iv(x,y)$ are harmonic by first developing the Cauchy-Riemann equations (CRE); $\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$.
Then, it is straightforward to show from the CRE that the real and imaginary parts are harmonic.
So, one can circumvent taking partial derivatives of $f$ by using the aforementioned theorems.
