5
$\begingroup$

This is a very basic question and I couldn't find it posted yet but here it goes; The Residue Theorem states that if $f:G\to \mathbb{C}$ is analytic on $G$- a region and $f$ has isolated singularities $b_1,...,b_k$ and $\gamma$ is homologous to 0, then $$\displaystyle\int_{\gamma}f=2\pi i\sum_1^kn(\gamma;b_s)\mathrm{Res}(f;b_s).$$ To my understanding, if $\gamma$ "wraps" around an isolated singularity, $b_i$, (say 1 time), then $\gamma$ would not be homologous to 0 since $n(\gamma;b_i)=1$ in this case ($b_i\in \mathbb{C}-G$). I think I need a better picture of the situation in this case. Thank you for your help!

$\endgroup$

2 Answers 2

4
$\begingroup$

To be homologous to zero, means we can deform the closed curve $\gamma$ to a point while staying in the region.

Taken from Tensor and Vector Analysis by C.E. Springer

Pretend $H$ is a singularity. Then $C_2$ can't be shrunk to a point instead it can be shrunk only to the boundary of $H$. If we shrink $C_2$ to a point (homologous to zero), we leave the region $K$. However, $C_1$ encircles no deleted neighborhoods or singularities so $C_1$ is homologous to zero.

Now the winding number is simply the number of times you circle the singularity.


Cauchy theorem states:

If $f(z)$ is analytic in $\Omega$, then $$ \int_{\gamma}f(z)dz = 0\tag{1} $$ for every cycle $\gamma$ which is homologous to zero in $\Omega$.

Residue theorem:

Let $f(z)$ be analytic except for isolated singularities $z_i$ in a region $\Omega$. Then $$ \frac{1}{2i\pi}\int_{\gamma}f(z)dz = \sum_in(\gamma,i)\operatorname{Res}\{f(z);z_i\}\tag{2} $$ for any cycle $\gamma$ which is homologous to zero in $\Omega$ and does not pass through and of the points $z_i$.

If every path in the residue theorem was homologous to zero, then equation $(1)$ would be equation $(2)$ and every integral you face would be zero. The winding number around the singularities will be $\pm 1$ depending on the direction of travel. If the curve is similar to $C_1$ in the image above, it is homologous to zero and we have equation $(1)$.

$\endgroup$
6
  • $\begingroup$ So in this picture, $C_2$ is not homologous to 0? Then for the Residue Theorem we couldn't use a curve such as $C_2$ but can only consider curves that are either $C_1$ or curves that wrap around $H$ in such a way so that the winding number is 0, correct? $\endgroup$
    – user23793
    Commented Mar 17, 2015 at 5:32
  • $\begingroup$ Can you deform $C_2$ to a point while staying in $K$? You can use the residue theorem when the contour is around a singularity. The residue theorem is $$\int_Cf(z)dz=2\pi i\sum\text{Res}$$ if there are no singularities the curve is homologous to zero so the right hand side is zero. $\endgroup$
    – dustin
    Commented Mar 17, 2015 at 5:38
  • $\begingroup$ Well in this picture the answer is no, you can't deform $C_2$ to a single point and still stay in $K$. $\endgroup$
    – user23793
    Commented Mar 17, 2015 at 5:41
  • $\begingroup$ So this was my question; we don't consider such curves as $C_2$ for the residue theorem because it does not satisfy the hypothesis. Correct? $\endgroup$
    – user23793
    Commented Mar 17, 2015 at 5:43
  • 15
    $\begingroup$ This answer confuses homologous with homotopic! $\endgroup$ Commented Dec 5, 2016 at 0:38
3
$\begingroup$

Your idea is mistaken. The fact that a curve has nonzero winding number around a single point gives no information about it being nullhomolgous. Rather, you want the following

A curve $\gamma$ in an open set $G$ of the plane is nullhomologous in $G$ if and only if $W(\gamma,z)=0$ for every $z\notin G$.

You can state the residue theorem as follows

Let $G$ be a region and $f:G\to \Bbb C$ be analytic except possibly at a finite set $A\subset G$. Let $\gamma:[0,1]\to G$ be a curve nullhomologous in $G$. Then $$\frac{1}{2\pi i}\int_\gamma f=\sum_{z\in A}W(\gamma,z)R(f,z)$$

$\endgroup$
5
  • $\begingroup$ I think I'm a little confused here. In this example $f=\frac{1}{z^2}$ and so is analytic on $\mathbb{C}-\{0\}$. Then if our contour is just the unit circle traversed once around this point, shouldn't the winding number be 1? $\endgroup$
    – user23793
    Commented Mar 17, 2015 at 5:22
  • $\begingroup$ @user23793 I had a typo, sorry. We're not calculating winding numbers alone, but also residues. The formula is $$\sum R(f,z)W(\gamma,z)$$ over certain points $z$. In this case $W(\gamma,0)=1$ but $R(f,0)=0$. At any rate, I have updated my answer to address your misconception. $\endgroup$
    – Pedro
    Commented Mar 17, 2015 at 5:55
  • $\begingroup$ In your definition of being homologous to zero, if there is a point outside of $G $ that isn't so that the winding number is zero, then wouldn't that mean the the curve does not meet the hypothesis of the RT? $\endgroup$
    – user23793
    Commented Mar 17, 2015 at 14:49
  • $\begingroup$ @PedroTamaroff: Are you familiar with a version of the theorem that allows the set $A$ to be discrete (i.e. in particular infinite)? How would one show that a for closed curve which is nullhomologous only a finite number of elements of $A$ have a nonzero winding number? $\endgroup$
    – el_tenedor
    Commented Mar 11, 2017 at 8:53
  • $\begingroup$ @el_tenedor Informally, note that the region determined by the curve is compact, so the set $A$ can only meet it in finitely many points, else it will accumulate. $\endgroup$
    – Pedro
    Commented Mar 11, 2017 at 19:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .