Solving Complex Analysis Problem: Calculating Index of a Curve

[Raziel2701]

Homework Statement

This is complex analysis by the way. Here's the problem statement:http://i.imgur.com/wegWj.png"

I'm doing part b, but some information from part a is carried over.

The Attempt at a Solution

My problem is that I don't know if I'm being asked to show it via direct substitution or if I should make use of the fact that the function is analytic on the disk. Or maybe if the integrand is supposed to simplify to the typical expression for the index (or winding number):

$$\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z-z_0}$$

If it is the case that I have to take advantage of the function being holomorphic, then wouldn't the integral be equal to zero though? Since f'(z) and f(z) are holomorphic(or analytic) in the disk?

Any thoughts?

 

[micromass]

In the integral

$$\int_\gamma{\frac{f^\prime(z)}{f(z)}dz}$$

What do you get when you substitute u=f(z)?

 

[Raziel2701]

I would get $$\int_\gamma \frac{1}{u}du$$ which would be 2pi i so all I have left is just one. So it helps, but I've yet to prove that it's the index. I'd expect a constant to be left.

I don't know if I have to incorporate the information relating to gamma and the composition f of gamma of t.

 

[micromass]

Not quite, if you make the substitution, you would end up with $$\int_{f\circ \gamma}{\frac{1}{u}du}$$ (you'll have to adjust the path to). So we have that

$$\frac{1}{2\pi i}\int_{\gamma}{\frac{f'(z)}{f(z)}dz}=\frac{1}{2\pi i}\int_{f\circ \gamma}{\frac{1}{z}dz}$$

But the right is the winding number of $$f\circ \gamma$$ around 0, which is what you had to show...

 

[Raziel2701]

Holy mackerel! So most of the information that was given in the problem was a distraction? Let me get this straight, we showed by substitution that the integrand is actually 1/u du, which allowed us to, well, show that the given expression is indeed the index number.

Did I get that right?

Lastly, for calculating the delta(t), I get $$e^{5it}$$ and $$e^{-3it}$$. Are these correct?

Thanks.

 

[micromass]

Yes... But the result you've proven is very important tho (even if its proof isn't very hard). It even has a name: the argument principle.

The delta(t) seems to be correct...