- #1
kuengb
- 106
- 0
I've got a homework problem from my function theory course that I never fully understood. It's getting hot again since the exams are nearing...please help
It goes simply like this: Compute the Integral
[tex]\int_{\vert z \vert=1}\frac{dz} {\sqrt{6 z^2-5 z +1}} [/tex]
The square root is chosen such that [itex] \sqrt{2}>0 [/itex]
One has to use that the integrand is holomorphic in C without the interval from 1/3 to 1/2 on the real line. But why is that? Why must one cut out the whole interval and not only the two "critical" points?
We didn't do much on square roots in class, the best I could find in my textbook was that a function that is holomorphic in a simply connected region + no zeroes does have a n-th root...I don't believe that's useful here.
Edit: I know how to actually solve the problem, that's why I didn't post this in the homework section. I just don't really understand the square root thing.
It goes simply like this: Compute the Integral
[tex]\int_{\vert z \vert=1}\frac{dz} {\sqrt{6 z^2-5 z +1}} [/tex]
The square root is chosen such that [itex] \sqrt{2}>0 [/itex]
One has to use that the integrand is holomorphic in C without the interval from 1/3 to 1/2 on the real line. But why is that? Why must one cut out the whole interval and not only the two "critical" points?
We didn't do much on square roots in class, the best I could find in my textbook was that a function that is holomorphic in a simply connected region + no zeroes does have a n-th root...I don't believe that's useful here.
Edit: I know how to actually solve the problem, that's why I didn't post this in the homework section. I just don't really understand the square root thing.
Last edited: