Contour Integral proof
Let be a polynomial of degree greater than 1, with all of it's zeroes inside the circle . show that .
My idea for a proof:
Since every polynomial over can be writen as , where are all the zeroes of (not necessarily unique). Than we can write .
Now we can say that for every we can find a circle such that
and therfore is analytic in so we get , but since we have and that shows what we needed to show.
I would appreciate some thoughts on this proof as I am not 100% sure it is valid.
consider the coordinate change , we have
Obviously are all the roots of , and for any k. That is all the roots of are outside of the circle ,
Since n>=2 the integrand is analytic inside the circle. DONE.
thanks for the reply, I found a hole in my proof as well so your direction is appreciated.
Thinking about this some more, what happens if the root is zero?
Here is an alternative:
Let . Since all of the roots are inside the circle , if , then
by Cauchy's integral theorem. (In other words, expanding the circle does not change the value of the integral.) Note that when is sufficiently large, then . Then, when is large,
(Keep in mind that since , we have .) Taking , the right hand side goes to zero, so the integral must be zero.
Thanks for the reply.
Originally Posted by roninpro
One question. Dosent needs to be analytic inside the circle to use Cauchys integral theorem. since the roots of are inside the circle it clearly isn't. Am I missing something here?
Another way to state my claim is that the value of the contour integral is zero in the region of the expansion (so that expanding the circle does not affect the original integral value). Since has no roots in that region, is analytic there, so it is okay to use Cauchy's integral theorem.
Originally Posted by skyking