Has $\frac{z}{\bar{z}}$ complex derivative at $z=0$? Has $\frac{z}{\bar{z}}$ complex derivative at $z=0$? I got zero from contour integral when contour is $z=\gamma(t)= e^{it}$, $t=[0,2\pi]$ so it should be analytic on unit circle.
 A: Have you noticed that
$$
\lim_{\substack{z \to 0 \\ z \in \mathbb{R}}} \frac{z}{\overline{z}} =1
$$
and
$$
\lim_{\substack{z \to 0 \\ z \in i\mathbb{R}}} \frac{z}{\overline{z}} =-1 ?
$$
The fact that the integral along a single, particular loop is zero has no relevance. There is Morera's theorem, but it deals with a continuous function defined in an open domain, and by assumption the integral along any loop must be zero.
A: The function $f(z):=z/\bar z$ is a priori undefined at $z:=0$; but maybe it can be defined there in a reasonable way, i.e., such that the extended $f$ is continuous there. So let's check this. Writing $z=re^{i\phi}$ with $r\geq0$ and $\phi\in{\mathbb R}$ we see that $$f(z)={re^{i\phi}\over  re^{-i\phi}}=e^{2i\phi}=\cos(2\phi)+i\sin(2\phi)\ ,$$ independently of $r\geq0$. This implies that $f$  assumes arbitrary values of absolute value $1$  in points $z$ arbitrarily close to $0$. It follows that the limit $\lim_{z\to0} f(z)$ does not exist; so there is no "reasonable way" to define $f(0)$. 
Since it is impossible to make $f$ continuous at $0$, a fortiori $f$ cannot be made differentiable there.
So let's forget about the origin. Maybe our $f$ is analytic in some region $\Omega\subset{\mathbb C}$ which does not contain the origin. If this were the case then the function $g(z):=z/ f(z)$, being the quotient of two analytic functions (with nonzero denominator), would also be analytic in $\Omega$. 
Now $g(z)=\bar z$. So let's test whether this has a complex derivative at some point $z_0\in\dot{\mathbb C}$. To this end we have to consider the limit
$$\lim_{h\to 0}{g(z_0+h)-g(z_0)\over h}=\lim_{h\to 0}{\overline{z_0+h}- \bar z_0\over h} =\lim_{h\to 0}{\bar h\over h}=\overline{\lim_{h\to 0}{ h\over \bar h}}\ .$$
As shown in the first part of the answer the limit on the right hand side does not exist. This proves that $g$, and whence $f$, does not have a complex derivative at any point $z_0\in\dot{\mathbb C}$. 
A: Differentiability implies continuity. But this functions is not continuous at $0$, even if you assign it a value there. Its limit along the real axis is $1$, and its limit along the imaginary axis is $-1$.
