MathematicalPhysicist
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In the pdf attached, I need help in questions:
4 and 5.
Here's what I did:
for 4 I had some thoughts how to tackle this but none were fruitful, one was to show that if they were any zeros in the interior of the curve, at least one then because f is analytic in this domain then this zero would be isolated, i.e there would be neighbourhood of the zero say a, s.t f(z) is different than zero (for every za<d, for some d>0), now somehow I need to show that eventually under this assumption we would get that Im(f(z)) on the contour is zero, well because f is anlaytic then the integral on the contour is zero, now because Im(f) on the contour is different than zero, it means that f doesnt equal zero on it, and if it had zero in the inetrior, then by the argument theorem
f(z)=e^(2pi*N) on the contour where N is the number of zeros (including their multiplicites), but that would mean that Im(f(z)) equals zero on the contour, correct or wrong yet again?
for those who track my posts know that I already asked on question 5 something like a month ago with no respond.
well after cheking on this again, I think that because f(0)=1 I need to compare via roche's theorem by comparing with the function: g(z)=f(z)1, but not sure how to do so, any ideas, obviously there's a simple trick here that I am overpassing here.
4 and 5.
Here's what I did:
for 4 I had some thoughts how to tackle this but none were fruitful, one was to show that if they were any zeros in the interior of the curve, at least one then because f is analytic in this domain then this zero would be isolated, i.e there would be neighbourhood of the zero say a, s.t f(z) is different than zero (for every za<d, for some d>0), now somehow I need to show that eventually under this assumption we would get that Im(f(z)) on the contour is zero, well because f is anlaytic then the integral on the contour is zero, now because Im(f) on the contour is different than zero, it means that f doesnt equal zero on it, and if it had zero in the inetrior, then by the argument theorem
f(z)=e^(2pi*N) on the contour where N is the number of zeros (including their multiplicites), but that would mean that Im(f(z)) equals zero on the contour, correct or wrong yet again?
for those who track my posts know that I already asked on question 5 something like a month ago with no respond.
well after cheking on this again, I think that because f(0)=1 I need to compare via roche's theorem by comparing with the function: g(z)=f(z)1, but not sure how to do so, any ideas, obviously there's a simple trick here that I am overpassing here.
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