TITLE: Showing a complex function is NOT holomorphic
QUESTION [6 upvotes]: I am a bit confused on all the different ways to show something is holomorphic, and I am wondering if my thoughts about the following are on the right track.

Show $$g(z)=\frac{z^*}{z^2+1}$$ is not holomorphic (where $z^{*}$ denotes
  the complex conjugate).

I know that a function is holomorphic iff its derivative with respect to $z^*$ is identically zero, but I am not sure I am taking the derivative with respect tot the conjugate correct.
Would it simply be $\frac{\partial g}{\partial z^{*}}=\frac{(z^2+1)(1)-(z^*)(0)}{(z^2+1)^2}=\frac{1}{z^2+1}$ which is only zero for $z=i$, so in general it isn't holomorphic? Is that correct reasoning or no?
And another example of a non holomorphic, $$h(z)=z((z^{*})^2-z^2)$$
$$=(zz^{*})z^{*}-z^2=|z|^{2}z^*-z^2$$
so would the $z^{*}$ derivative just be $|z|^2$ which is zero only at the origin? I'm just confused on how to actually show this. I think I am also confused on taking the deravtive with respect to the conjugate.
I know I could also trying separating into the form of $U(x,y)+IV(x,y)$ and showing that the CR don't hold, but that seems even more difficult. Looking for any and all help/advice. Thank you

REPLY [2 votes]: Yes, you are correct. As you mention a complex function $f$ is holomorphic on some open set $G \subset \mathbb{C}$ iff 
\begin{align}
\frac{\partial f}{\partial \bar z}  = 0.
\end{align} 
Note we also have that
\begin{align}
\frac{\partial}{\partial \bar z} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y} \right).
\end{align}
which means
\begin{align}
\frac{\partial g}{\partial \bar z} =&\ \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y} \right) \frac{x-iy}{x^2-y^2+1+i2xy}\\
=&\ \frac{(x^2-y^2+1+i2xy)-(x-iy)(2x+i2y)}{2(x^2-y^2+1+i2xy)^2}+i  \frac{-i(x^2-y^2+1+i2xy)-(x-iy)(-2y+i2x)}{2(x^2-y^2+1+i2xy)^2}\\
=& \frac{(z^2+1)}{(z^2+1)^2} = \frac{1}{z^2+1}
\end{align}
{"set_name": "stack_exchange", "score": 6, "question_id": 1937632}