- #1
sari
- 23
- 1
1. find the number of solutions of e^iz - z^2n - a = 0 in the upper half of the complex plane, where n is a natural number and a is a real number such that a>1.
2. Rouche's theorem: If f and g are analytic functions in a domain, and |f|>|g| on the boundary of the domain, then the number of zeros of f+g within the domain equals the number of zeros of f within the domain (including multiplicities).
i tried breaking e^iz - z^2n - a into two functions f and g, and showing that one is greater than the other along the upper part of a closed half-circle of radius r around the origin (i.e, [-r,r] + {z: |z|= r, z>0}).
for any z with Imz >= 0 (in the upper half of the plane),
|e^iz| = e^-Imz <= 1.
Along the contour, |z^2n|<= r^2n
and |a|>1.
I tried using the triangle inequality (||a|-|b|| <= | a+b | <= |a|+|b| ) with the three possible combinations of functions:
1. e^iz - z^2n ; a
2. e^iz ; - z^2n - a
3. e^iz - a ; - z^2n
yet I did not succeed in finding a combination such that one function would always be larger than the other along the chosen contour.
Except, if r^2n < a-1 then we can use combination 1 and get that | e^iz - z^2n | < 1+a-1=a, so e^iz - z^2n - a has the same number of zeros as a (none). But that only helps for a tiny part of the upper plane.
Then I tried finding a contour that was sort of a "half ring" (with "?"< r_1<r_2) over which I could evaluate the functions (of course, r_2 can technically be as big as we like, so that takes care of {z: Imz > 0}\{z: |z|<=r_1}, but that didn't seem to work either (even supposing I could prove that there are no zeros in |z|<=r_1).
I thought I could somehow make use of the fact that - z^2n - a has no zeros along the real line.
I also thought that maybe I should use a different shaped contour (like a rectangular contour), but I can't really see how that would help. Or I could try to use a circle of radius r (which can be as big as we want) around the origin, and somehow show that all the zeros are in the upper half of the plane. Though it doesn't really make sense to do that, because then you lose the bound on e^iz.
I also thought that the half-circle contour reminded me of how we learned how to use the residue theorem to calculate improper integrals of real functions (from -infty to infty) by integrating along half-circles of as large a radius as we wish and taking the limit as r goes to infinity, but aside from the similar outer appearance of the contours, I can't see how that would be applicable in this case.
Additional things we know that could be useful:
- z^2n - a has 2n zeros along |z|=a^1/2n. Because of the symmetry of the roots, half of them (n) are in the upper half of the plane.
e^iz - a has no zeros.
I would appreciate any ideas! I feel the answer is sitting right in front of me and just not "clicking".
Thanks!
2. Rouche's theorem: If f and g are analytic functions in a domain, and |f|>|g| on the boundary of the domain, then the number of zeros of f+g within the domain equals the number of zeros of f within the domain (including multiplicities).
The Attempt at a Solution
i tried breaking e^iz - z^2n - a into two functions f and g, and showing that one is greater than the other along the upper part of a closed half-circle of radius r around the origin (i.e, [-r,r] + {z: |z|= r, z>0}).
for any z with Imz >= 0 (in the upper half of the plane),
|e^iz| = e^-Imz <= 1.
Along the contour, |z^2n|<= r^2n
and |a|>1.
I tried using the triangle inequality (||a|-|b|| <= | a+b | <= |a|+|b| ) with the three possible combinations of functions:
1. e^iz - z^2n ; a
2. e^iz ; - z^2n - a
3. e^iz - a ; - z^2n
yet I did not succeed in finding a combination such that one function would always be larger than the other along the chosen contour.
Except, if r^2n < a-1 then we can use combination 1 and get that | e^iz - z^2n | < 1+a-1=a, so e^iz - z^2n - a has the same number of zeros as a (none). But that only helps for a tiny part of the upper plane.
Then I tried finding a contour that was sort of a "half ring" (with "?"< r_1<r_2) over which I could evaluate the functions (of course, r_2 can technically be as big as we like, so that takes care of {z: Imz > 0}\{z: |z|<=r_1}, but that didn't seem to work either (even supposing I could prove that there are no zeros in |z|<=r_1).
I thought I could somehow make use of the fact that - z^2n - a has no zeros along the real line.
I also thought that maybe I should use a different shaped contour (like a rectangular contour), but I can't really see how that would help. Or I could try to use a circle of radius r (which can be as big as we want) around the origin, and somehow show that all the zeros are in the upper half of the plane. Though it doesn't really make sense to do that, because then you lose the bound on e^iz.
I also thought that the half-circle contour reminded me of how we learned how to use the residue theorem to calculate improper integrals of real functions (from -infty to infty) by integrating along half-circles of as large a radius as we wish and taking the limit as r goes to infinity, but aside from the similar outer appearance of the contours, I can't see how that would be applicable in this case.
Additional things we know that could be useful:
- z^2n - a has 2n zeros along |z|=a^1/2n. Because of the symmetry of the roots, half of them (n) are in the upper half of the plane.
e^iz - a has no zeros.
I would appreciate any ideas! I feel the answer is sitting right in front of me and just not "clicking".
Thanks!