Show three ways that $f(z)=\frac{\overline{z}}{z-1}$ is not analytic I need to show the complex function $$f(z)=\frac{\overline{z}}{z-1}$$ is not analytic in three ways; using Cauchy's equations, geometrically, and by integrating over the circle of radius 2. 
I used Cauchy's equations using $u(x,y)=\frac{x^2-x-y^2}{(x-1)^2+y}$ and $v(x,y)=\frac{y-2xy}{(x-1)^2+y}$.The calculations were complicated, but I managed to show they were not equal.
For the second part, I chose the lines $x=1$ and $y=1$, mapped them, and showed that the slopes of their tangents at their intersection points did not meet at a right angle. I believe this answers the question.
My biggest question lies in the integral. I did it, but got 0. Here is what I did:
We want $$\oint_C \frac{\overline{z}}{z-1}$$, where $C$ is given by $|z|=2$. I chose to parameterize $z$, giving $z=2e^{it}$ for $0\leq t\leq2\pi$. This gives $\overline{z}=2e^{-it}$ and $dz=2ie^{it}dt$, and so I integrated the following: $$\int_0^{2\pi} \frac{2e^{-it}}{2e^{it}-1}2ie^{it}dt.$$ Unfortunately, upon integration, I got that the integral is equal to $0$. I didn't expect this, as the integral should vanish if it is analytic, but it clearly is not at $z=1$. Did I make a mistake?
Thanks!
 A: I don't believe you made a mistake at all; in fact, what you found is in line with what you should expect.
When you are asked to show that $f(z)$ is not analytic, they mean other than at $z = 1$, where it is clearly not analytic since it his a pole. However, if it were analytic other than at $z = 1$, you would be able to use the residue theorem (since there is a pole inside the circle) to conclude that
$$
\int_\Gamma f(z)dz = 2\pi i Res_{z=1}f(z) = 2\pi i
$$
However, you found that the integral was zero. So the function is not analytic (away from the pole).
A: Concerning the difficulty of the Cauchy-Riemann equation approach: if $f(z)$ were analytic anywhere except at the point $z=1$, then $(z-1)f(z)$ would also be analytic. But $(z-1)f(z)=\bar{z}$ is not analytic anywhere, since $\frac{\partial{\bar{z}}}{\partial{x}}+i\frac{\partial{\bar{z}}}{\partial{y}}=1+i(-i)=2\neq 0$.
In general, if $f(z)=\bar{z}g(z)$, where $g(z)\neq 0$ is analytic, then $\frac{\partial{f}}{\partial{x}}=\bar{z}\frac{\partial{g}}{\partial{x}}+g(z)$ and $\frac{\partial{f}}{\partial{y}}=\bar{z}\frac{\partial{g}}{\partial{y}}-ig(z)$. It follows that, $\frac{\partial{f}}{\partial{x}}+i\frac{\partial{f}}{\partial{y}}=\bar{z}\left (\frac{\partial{g}}{\partial{x}}+i\frac{\partial{g}}{\partial{y}}\right )+2g(z)=2g(z)\neq 0$. Therefore, the Cauchy-Riemann equation is not satisfied and the function is not analytic anywhere in the complex plane.
