Complex Analysis (Argument Principle to determine location of roots)

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Homework Statement


With [tex]f(z) = 2z^{4} +2z^{3} +z^{2} +8z +1[/tex]

Show that [tex]f[/tex] has exactly one zero in the open first quadrant.


Homework Equations


Argument Principle


The Attempt at a Solution


I know I'm supposed to use the Argument Principle.. So far, all I can do is show something like, in the unit disc there exists ONE zero (by Rouche's Theorem, using function [tex]8z+1[/tex] as my upper bound.

However, showing something in a specific Quadrant is proving to be more difficult. I thought about using the same function [tex]g(z) = 8z+1[/tex] and to say that since [tex]|f(z) - g(z)| \leq |g(z)|[/tex] in that quadrant, then it is proved but obviously that's not rigorous.

So I tried defining [tex]h(z) = f(z) - g(z) = 2z^{4} +2z^{3} +z^{2}[/tex] and define the set [tex]Q = { z = x+iy : x,y>0 } [/tex] and say that for sufficiently large [tex]|z|, z \in Q [/tex] we have that [tex]|g(z)|/|h(z)| \leq 1[/tex] since the LHS in fact equals zero, since it grows faster.

I'm pretty sure that's not good enough.. And I'm not utilizing the Argument Principle, can someone help? Thanks!
 

Answers and Replies

  • #2
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Woo, another theorem in complex analysis that we apparently proved but did not use. Can you just consider the quarter-circle in the first quadrant of arbitrary radius R > 0 (so the real axis from 0 to R, then Re^(it) for t between 0 and pi/2, and then come down the imaginary axis from Ri to 0)? Then finding the winding number about f(z) = 0 of this curve might work.
 

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