If a function g(z) is nowhere differentiable, then is f(g(z)) also nowhere differentiable? I have the function $\sin(\bar{z})$, and have shown (I think) that $g(z)=\bar{z}$ is differentiable nowhere. Does this mean that $\sin(\bar{z})$ is also not differentiable?
 A: In general no, though (regarding the question in the title).  For an easy counterexample: take $f(z)=g(z)=\bar z$. Then $f(g(z))=z$.
For your particular example, $\sin \bar z$ is not $\mathbb{C}$-differentiable at most points. However, with $z=x+iy$,
$$
\sin \bar z = \sin(x-iy) = \sin x \cosh y - i \cos x \sinh y = u + iv.
$$
Hence
\begin{align}
u'_x &= \cos x \cosh y & u'_y &= \sin x \sinh y \\
v'_x &= \sin x \sinh y & v'_y &= -\cos x \cosh y
\end{align}
which shows that Cauchy-Riemann's equations are satisfied at points where
$\cos x = 0$ and $\sinh y = 0$, i.e. at points where $z = \pm \frac\pi2 + 2\pi k$. (These are exactly the points where $f'(z) = 0$.) Since $u$ and $v$ are $C^1$, $\sin \bar z$ is in fact differentiable at these points.
I'll leave it to you to generalize the above idea (easiest via Wirtinger derivatives).
A: I think so, if f itself is holomorphic; assume not, i.e., assume $sin (\bar {z})$ is holomorphic in some open set $U$, then consider a local inverse $sin^{-1}$ for $sinz$ overlapping $U$, which is holomorphic by the inverse function theorem , and then in the overlap, you have the composition of two holomorphic functions $sinz, sin^{-1}z $ which is holomorphic, but agrees with the nowhere-holomorphic conjugation map $\bar{z}$. This argument extends to any holomorphic function $g(z)$ composed with $\bar{z}$
A: Take some pathologic function like:
$$
f(x) = \begin{cases}1 & \text{$x$ rational} \\
                    0 & \text{otherwise}
       \end{cases}
$$
Then $f$ is nowhere continuous, thus nowhere differentiable. Same for $g(x) = \sqrt(2) (1 + f(x))$, which is irrational everywhere. Then $f(g(x)) = 0$, which is differentiable.
