A problem on Residue Theorem Today I had a problem in my test which said

Calculate $\int_C \dfrac{z}{z^2 + 1}$ where C is circle $|z+\dfrac{1}{z}|= 2$.

Now, clearly this was a misprint since C is not a circle. I tried to find the 'curve' C, by converting to Cartesian, but I found that C is set of isolated points in the complex plane. During test, I could only find two points which satisfy the curve, $z=1 $ and $z= -1$. However, I came home and searched on wolfram alpha that the 'curve' is actually set of 6 isolated points. So my question is, since the poles of the function to be integrated i.e. $\dfrac{z}{z^2+1}$ are 'outside' the curve i.e. $\pm i$ do not satisfy the 'curve', is the value of integral by residue theorem $0$?
If yes or no, what are the explanations?
 A: The answer is no (in my opinion), because the  residue theorem is only true on domains, still the value of your integral is $0$ as isolated points are so called null sets, which means that the integral over them vanishes anyway, no matter what function you are integrating.
In fact for the Residue theorem you need the function to be holomorphic in the domain (with exception of the poles), and defining holomorphic on singletons should be difficult.
I guess this question leads to a discussion.
A: I'm almost certain that your question was incorrectly stated. However, the "curve" in the question is the following:
$$
\left| z + \frac1z \right| = 2 \Leftrightarrow 
\left| z + \frac1z \right|^2 = 4 \Leftrightarrow 
| z^2 + 1 |^2 = 4|z|^2 \Leftrightarrow
| z^2 + 1 |^2 - 4|z|^2 = 0
$$
Put $z=x+iy$, then
\begin{align}
| z^2 + 1 |^2 - 4|z|^2 &= x^4 +2x^2y^2 -2x^2 + 1 - 6y^2 + y^4 \\
&= (x^2-1+2y+y^2)(x^2-1-2y+y^2) \\
&= (x^2 + (y+1)^2 - 2)(x^2 + (y-1)^2-2)
\end{align}
almost as by magic.
So, your curve consists of two circles, $|z+i| = \sqrt2$ and $|z-i|=-\sqrt2$.

I leave it up to you to guess exactly which curve the problem constructor wanted you to integrate over.
