# Complex line integrals

PROBLEM:

Show that the integral $$\int_{C}^{}\frac{dz}{z^{2}}$$ 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 $$\frac{1}{z^{2}}$$ 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:

$$z=re^{i\theta}$$ for any circle with radius r.
$$dz=ire^{i\theta}d\theta$$
-a will become $$\pi$$. b will become 0.

$$\int_{\pi}^{0}\frac{ire^{i\theta}}{re^{i\theta}re^{i\theta}}d\theta$$

$$= \int_{\pi}^{0}\frac{i}{r}e^{-i\theta}d\theta$$

$$=-\frac{1}{r}\int_{\pi}^{0}-ie^{-i\theta}d\theta$$

$$= -\frac{1}{r}(e^{-i\theta})^{0}_{\pi}$$

$$= -\frac{2}{r}$$

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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## Answers and Replies

Avodyne
Science Advisor
I know from a past assignment that $$\frac{1}{z^{2}}$$ has a branch point at the origin.

Actually, it's a pole, not a branch point.

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.

Yes!

(B.) The path as the same length regardless of your choice of a and b.

No!

how can I show that ANY path will have the same length?

The length of the path is not relevant. The integral is independent of all aspects of the path other than its starting and ending points.

The length of the path is not relevant. The integral is independent of all aspects of the path other than its starting and ending points.

Dear god, I don't even know what these integrals are evaluating. I'm going to search the web a bit.

I have a new idea, though for this question... I'll say that f(z) is continuos and analytic except at z = 0. F'(z) = f(z) F(z) = -x^-1 is also continuous and analytic. So, as long as I avoid the origin, I can use the fundamental theorem of calculus to evaluate the integral. With a and b both constant it only gives one result for that a and b.

Dick
Science Advisor
Homework Helper
Take two paths P1 and P2 between a and b. Put together they make a closed path C. If the origin is not enclosed in C then f is analytic in the interior of C, what do you conclude about the integral of f around C? What does this tell you about the integrals along P1 and P2? If the origin is in C then you effectively have a closed integral around the origin of 1/z^2. You reach the same conclusion for a slightly different reason.

Dick
Science Advisor
Homework Helper
PS. Not going through the origin doesn't have anything to do with branch points here. 1/z^2 is not multiple valued and doesn't need any branch definitions. But the origin is still a singularity.