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[complex functions] finding complex roots in 1+z+az^n

  • Thread starter rahl___
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  • #1
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Hi,

I have a big problem in solving such question:

Let [tex]W(z) = 1 + z + az^n[/tex], where [tex]a[/tex] is complex and [tex]n[/tex] is natural and greater than 1. Show that [tex]W(z)[/tex] has a root that satisfies [tex]|z_k| <=2 [/tex].
I have no ideas how to solve it. I thought about integrating W and showing that it's roots create a circle with radius equal to 2, but it completely didnt work. I would appreciate if someone could give me a clue, as I really cant see any way of solving this one.

rahl
 

Answers and Replies

  • #2
HallsofIvy
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Well, of course, the roots do NOT "create a circle with radius equal to 2".


Show that the absolute value, |W(z)|, is greater than some positive value for |z|> 2, so that there is no solution outside that circle.
 
  • #3
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of course they dont, I dont know what I was thinking when writing that.

Show that the absolute value, |W(z)|, is greater than some positive value for |z|> 2, so that there is no solution outside that circle.
I dont understand what you are implying.
Let [tex]z=4[/tex] and [tex]a = \frac{-5}{4^n}[/tex]. In that case [tex]|W(4)|=0[/tex], so there obviously is a solution outside that circle.

EDIT:
I have googled a hint:
http://www.math.cornell.edu/~barbasch/courses/418-06/p2sz.pdf
problem 3 is exactly my problem. Do I have to know what do these two polynomials W(z) and W(1/z) have in common, to succesfully use that replacement? I realize that W(1/z) will have the same roots as [tex](z^n+z^{n-1}+a)[/tex], which is a polynomial that we can get, when switching the coefficients in W(z): the first one with the last one, the second one with the prelast one, and so on. But what do they have in common apart from switched coefficients? And what about the roots of that switched polynomial? Where are they situated? I cant find any answer to this questions, I'd be grateful for any clues.

EDIT2:
I've finally solved it. Moderators can delete this thread or I can write the solution and let the thread live.

rahl
 
Last edited:

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