# 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?

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I'm sure $C$ is note a discrete set of point – Selim Ghazouani Mar 11 '13 at 9:30
As in? Then what does C represent? – taninamdar Mar 11 '13 at 10:20

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.

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Whoa! I had reached till the Cartesian conversion, however had no clue how to factorize it or whether it's factorizable. Anyway, thanks for answering. Now it depends how the problem setter wanted us to integrate over, as one can choose the orientation of countor in such a way that both poles are inside or outside C. – taninamdar Mar 11 '13 at 11:25

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.

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Well, if residue theorem cannot be applied because of the constraints, we can say that the value is $0$ by Cauchy's Theorem, right? – taninamdar Mar 11 '13 at 7:37
no for cauchys theorem you need the bound of a domain too, and your set cannot be written as the bound of a domain, it is $0$ because your set is finite – Dominic Michaelis Mar 11 '13 at 7:39
I see. Still I would like to hear explanation from more people. :) – taninamdar Mar 11 '13 at 7:40
$\left|\int_{E}f(z)\text{d}z \right| \leq \mathbf{m}(E) \sup|f(z)|$ – Laura Mar 11 '13 at 7:45
Can you explain what you mean by that $m(E) \sup |f(z)|$? I am an engineering student so I am not so familiar with the terminology. – taninamdar Mar 11 '13 at 10:19