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Apr
1
awarded  Notable Question
Mar
17
comment Residue Theorem and Homologous to zero
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?
Mar
17
comment Residue Theorem and Homologous to zero
This is the heart of my question I think. If $C_2$ is NOT homologous to 0, then it does not satisfy the hypothesis because in order to use the Residue Theorem, we need a curve that is homologous to 0. I apologize if I'm being stupid here...
Mar
17
comment Residue Theorem and Homologous to zero
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?
Mar
17
comment Residue Theorem and Homologous to zero
Well in this picture the answer is no, you can't deform $C_2$ to a single point and still stay in $K$.
Mar
17
comment Residue Theorem and Homologous to zero
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?
Mar
17
comment Residue Theorem and Homologous to zero
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?
Mar
17
asked Residue Theorem and Homologous to zero
Mar
1
accepted Morera's Theorem and annuli
Mar
1
answered Morera's Theorem and annuli
Mar
1
asked Morera's Theorem and annuli
Feb
19
comment Consider the function
I would try a sequential argument and use the denseness of $\mathbb{R}$.
Feb
19
comment A Step in the Proof of Green's Theorem
Which limits are you talking about? Could you be more specific?
Feb
17
revised Did I do this proof right?
Here is one proof.
Feb
12
awarded  Teacher
Feb
12
answered Did I do this proof right?
Feb
1
awarded  Critic
Feb
1
comment If $x_n \rightarrow 0$ and $\{y_n\}$ is a bounded sequence, then $x_ny_n \rightarrow 0$.
Just be careful of your $\alpha$. It must be a point so that $\alpha \in \mathbb{R}^+$. And yes, the same $N$ works because $y_n$ is bounded independent of whether $x_n$ converges or not.
Jan
12
accepted Part of Fubini's Theorem with almost everywhere
Jan
12
comment Part of Fubini's Theorem with almost everywhere
This question was already asked math.stackexchange.com/questions/1092685/…