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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!