What does it mean when the second Wirtinger derivative is sometimes zero? I have to show that $f(z) =\sin (\bar z)$ is not analytic anywhere.
One way is to check the CR equations by letting $\sin (z) = \sin(x+iy)$ and do some algebra. From the CR equations I obtained that $u_x = v_y$ occurs when $\cos(x) = 0$ and $\cosh(y)=0$. On the other hand, $v_x = -u_y$ occurs when $\sin(x)=0$ and $\sinh(y)=0$. The intersection of both sets of solutions is the empty set and thus I conclude $f(z)$ is not analytic anywhere.
But I figured why not try the Wirtinger derivatives since the function clearly is defined in terms of $\bar z$. So doing the partial derivative with respect to $\bar z$ I obtain the expression $\frac{\partial f}{\partial \bar z} = \cos(\bar z)$, which, while not identically zero, can be zero at certain values of $\bar z$.
My question is, I know that $\frac{\partial f}{\partial \bar z} = 0$ if and only if $f$ is analytic, but what if $\frac{\partial f}{\partial \bar z}$ is only zero at certain points? Does it mean it is not analytic anywhere? Does it mean it is analytic at those points? Does it even matter that $\frac{\partial f}{\partial \bar z}$ is an expression that can sometimes be zero?
I did consider using examples with singular points, like $f(z)=\frac{1}{z}$ to try and figure out how $\frac{\partial f}{\partial \bar z}$ looks like for a function that is analytic somewhere but not everywhere but in this case $\frac{\partial f}{\partial \bar z}=0$, and one example is never a deal breaker.
I know it doesn't really make sense to talk about possible analyticity on a point since a point is not an open set, but really I can't understand it.
 A: $\newcommand{\dd}{\partial}$The map $z\mapsto\sin\bar z$ is holomorphic (at a point) sometimes. You should know that the Wirtinger derivative in the conjugate $\bar z$ is zero at a point $z_0$ if and only if the Cauchy-Riemann equations hold at that point $z_0$ (take real and imaginary parts on the definition).
Let's use the Wirtinger derivatives from their definition:

$$\frac{\dd}{\dd z}=\frac{1}{2}\left(\frac{\dd}{\dd x}-i\frac{\dd}{\dd y}\right),\,\frac{\dd}{\dd\bar z}=\frac{1}{2}\left(\frac{\dd}{\dd x}+i\frac{\dd}{\dd y}\right)$$

Application to $(x+iy)\overset{\sin\bar z}{\mapsto}\sin(x-iy)$ gives: $$\begin{align}\frac{\dd}{\dd\bar z}\sin\bar z&=\frac{1}{2}\left(\frac{\dd}{\dd x}\sin(x-iy)+i\frac{\dd}{\dd y}\sin(x-iy)\right)\\&=\frac{1}{2}\left(\cos(x-iy)+i(-i)\cos(x-iy)\right)\\&=\cos(x-iy)\end{align}$$
Exactly as you say, which equals zero if and only if $z=x+iy=\pi(n+1/2),\,n\in\Bbb Z$. What about the Cauchy Riemann equations? Let's be careful with what the equations imply.

$$\begin{align}\sin\bar z&=\sin(x)\cosh(y)-i\cos(x)\sinh(y)\\\\\frac{\dd u}{\dd x}&=\frac{\dd}{\dd x}\sin(x)\cosh(y)=\cos(x)\cosh(y)\\\frac{\dd v}{\dd y}&=\frac{\dd}{\dd y}(-\cos(x)\sinh(y))=-\cos(x)\cosh(y)\\&=\frac{\dd u}{\dd x}\iff0\in\{\cos(x),\cosh(y)\}\\\\\frac{\dd u}{\dd y}&=\frac{\dd}{\dd y}\sin(x)\cosh(y)=\sin(x)\sinh(y)\\\frac{\dd v}{\dd x}&=\frac{\dd}{\dd x}(-\cos(x)\sinh(y))=\sin(x)\sinh(y)\\&=-\frac{\dd u}{\dd y}\iff0\in\{\sin(x),\sinh(y)\}\end{align}$$

Can we reconcile the two? Yes, we can. It seems as if all your calculations were correct, but your deductions had a mistake. First of all, $0\in\{\cos(x),\cosh(y)\}$ iff. $\cos(x)=0$ since $\cosh$ is nonzero on the real line. So the C-R criterion for differentiability (which is in fact identical to the Wirtinger criterion) tells us that our function is differentiable iff. $\cos(x)=0$ and one or both of $\sin(x),\sinh(y)$ are zero. However, $\sin^2(x)=1-\cos^2(x)=1$ if $\cos(x)=0$, so we have differentiability iff. $\cos(x)=0$ and $\sinh(y)=0$. The $\sinh$ function has only one real root, namely at $y=0$. Then, CR is stating that $y=0$ and $x$ is such that $\cos(x)=0$, so $z=\pi(n+1/2),\,n\in\Bbb Z$ exactly as the Wirtinger derivative criterion (much more efficiently) computed.
There are no contradictions here. More generally, if $f$ is holomorphic then $z\mapsto f(\bar z)$ is holomorphic at $z_0$ iff. $f'(z_0)=0$ (this is consistent with our example - at those mentioned $z$, $(\sin z)'=\cos(z)=0$). Try to prove this using the Wirtinger criterion.
