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futurebird
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PROBLEM:
Show that the integral [tex]\int_{C}^{}\frac{dz}{z^{2}}[/tex] where C is a path beginning at z=-a and ending a z=b, where a > 0 and b >0, is independent of path so long as C doesn't go through the origin.
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WHAT I HAVE DONE:
I know from a past assignment that [tex]\frac{1}{z^{2}}[/tex] has a branch point at the origin. That is why the path C can't go through the origin. I imagine the start point, -a on the negative x-axis (since they have set it up so -a is negative and real) and the end point, b, on the positive x-axis. The path from -a to be is shaped like a rainbow or semi-circle to avoid the origin. The path could have turns in it, but it won't intersect itself.
By "independent of path" do they mean that:
(A.) Given a and b, you can take any path from -a to b and get the same result for that specific a and b. That is, if a=2 and b=30 there is one answer with many paths from -2 to 30. But, if you have a=0.5 and b=77 the answer could be different, although you still have many paths between -a and b.
(B.) The path as the same length regardless of your choice of a and b.
I hope it is not "B." because if:
[tex]z=re^{i\theta}[/tex] for any circle with radius r.
[tex]dz=ire^{i\theta}d\theta[/tex]
-a will become [tex]\pi[/tex]. b will become 0.
[tex]\int_{\pi}^{0}\frac{ire^{i\theta}}{re^{i\theta}re^{i\theta}}d\theta[/tex]
[tex]= \int_{\pi}^{0}\frac{i}{r}e^{-i\theta}d\theta[/tex]
[tex]=-\frac{1}{r}\int_{\pi}^{0}-ie^{-i\theta}d\theta[/tex]
[tex]= -\frac{1}{r}(e^{-i\theta})^{0}_{\pi}[/tex]
[tex]= -\frac{2}{r}[/tex]
What this says is that, if the path is a semi-circle, then the length is dependent on r. But this would only work when a = b, still I think it shows that the vale of the integral chages depending on the values of a and b. So they can't be asking that I show that: "The path as the same length regardless of your choice of a and b."
I need to know if I understand the question correctly, and, how can I show that ANY path will have the same length?
Show that the integral [tex]\int_{C}^{}\frac{dz}{z^{2}}[/tex] where C is a path beginning at z=-a and ending a z=b, where a > 0 and b >0, is independent of path so long as C doesn't go through the origin.
-----------
WHAT I HAVE DONE:
I know from a past assignment that [tex]\frac{1}{z^{2}}[/tex] has a branch point at the origin. That is why the path C can't go through the origin. I imagine the start point, -a on the negative x-axis (since they have set it up so -a is negative and real) and the end point, b, on the positive x-axis. The path from -a to be is shaped like a rainbow or semi-circle to avoid the origin. The path could have turns in it, but it won't intersect itself.
By "independent of path" do they mean that:
(A.) Given a and b, you can take any path from -a to b and get the same result for that specific a and b. That is, if a=2 and b=30 there is one answer with many paths from -2 to 30. But, if you have a=0.5 and b=77 the answer could be different, although you still have many paths between -a and b.
(B.) The path as the same length regardless of your choice of a and b.
I hope it is not "B." because if:
[tex]z=re^{i\theta}[/tex] for any circle with radius r.
[tex]dz=ire^{i\theta}d\theta[/tex]
-a will become [tex]\pi[/tex]. b will become 0.
[tex]\int_{\pi}^{0}\frac{ire^{i\theta}}{re^{i\theta}re^{i\theta}}d\theta[/tex]
[tex]= \int_{\pi}^{0}\frac{i}{r}e^{-i\theta}d\theta[/tex]
[tex]=-\frac{1}{r}\int_{\pi}^{0}-ie^{-i\theta}d\theta[/tex]
[tex]= -\frac{1}{r}(e^{-i\theta})^{0}_{\pi}[/tex]
[tex]= -\frac{2}{r}[/tex]
What this says is that, if the path is a semi-circle, then the length is dependent on r. But this would only work when a = b, still I think it shows that the vale of the integral chages depending on the values of a and b. So they can't be asking that I show that: "The path as the same length regardless of your choice of a and b."
I need to know if I understand the question correctly, and, how can I show that ANY path will have the same length?
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