I am having a little bit of trouble with the following:

$$\int_{\gamma}\frac{z^2-1}{z^2+1}dz$$ where $\gamma$ is a circle of radius $2$ centered at 0. I am trying to separate this or simplify it into the form in which we can maybe apply Cauchy's differentiation formula but this isn't working. I did get this:

$$\frac{z^2-1}{z^2+1} = 1-\frac{2}{z^2+1}$$ but this doesn't seem to take me anywhere. Any hints would be helpful.


Would it be possible to do this:

$$\frac{z^2-1}{z^2+1} = 1-\frac{2}{z^2+1}= 1-\frac{\frac{2}{z+i}}{z-i}$$ and then just apply cauchy's integral formula with $z_0=i$? and $f(z)=\frac{2}{z+i}$?

Using this, we would have

$$ \frac{2}{i+i}2\pi i + \frac{2}{-i-i}2\pi i = 0 $$ as the answer?

  • $\begingroup$ You have another pole, at $-i$, that is inside the circle. ${}\qquad{}$ $\endgroup$ – Michael Hardy Mar 9 '15 at 7:10
  • $\begingroup$ @MichaelHardy My trouble is I don't particularly understand what you mean by pole. I'm assuming you mean a point in which we have a problem as in $i$ and $-i$ because this gives us the two trouble points inside the curve. I don't know where to go from here. I just learned about Cauchy's Integral Formula. $\endgroup$ – H5159 Mar 9 '15 at 7:11
  • $\begingroup$ @MichaelHardy Would we do two different integrals one for $-i$ and one for $i$ and add them together? Should the answer be 0? $\endgroup$ – H5159 Mar 9 '15 at 7:15
  • $\begingroup$ I posted a method using partial fraction expansion, which leads to a fairly quick result that might be surprising. $\endgroup$ – Mark Viola Mar 9 '15 at 16:41
  • $\begingroup$ I guess the problem is that Frumpy knows a "Cauchy differentiation formula" but doesn't know the "Cauchy integral formula". And doesn't know what is a pole, so certainly doesn't know what is a residue. $\endgroup$ – GEdgar Mar 9 '15 at 16:50

$$ \int_\gamma 1\,dz=0 $$ because $\gamma$ returns to its starting point.

Next we have $$ z^2+1 = (z-i)(z+i). $$

So the integral should involve the sum of residues at $\pm i$, since $\gamma$ winds once around each of those two points.

PS: Your proposal to apply Cauchy's formula at $i$ to the function $$ 1-\frac{\frac{2}{z+i}}{z-i} $$ would work if not for the fact that the numerator $\dfrac2{z+i}$ also has a pole inside the curve $\gamma$. You need to take the residue at that point into account as well.

  • $\begingroup$ I made an edit, could you check out my work? $\endgroup$ – H5159 Mar 9 '15 at 7:05
  • $\begingroup$ I have done some more work in the OP. $\endgroup$ – H5159 Mar 9 '15 at 7:21

Partial fraction expansion yields $$\frac{z^2-1}{z^2+1}=\frac{z^2-1}{2i}\left(\frac{1}{z-i}-\frac{1}{z+i}\right)$$ and application of the Residue Theorem reveals that the integral is $$=2\pi i\left(\frac{(i)^2-1}{2i}-\frac{(-i)^2-1}{2i}\right)=0$$


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