# Complex Analysis (Argument Principle to determine location of roots)

## Homework Statement

With $$f(z) = 2z^{4} +2z^{3} +z^{2} +8z +1$$

Show that $$f$$ 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 $$8z+1$$ as my upper bound.

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

So I tried defining $$h(z) = f(z) - g(z) = 2z^{4} +2z^{3} +z^{2}$$ and define the set $$Q = { z = x+iy : x,y>0 }$$ and say that for sufficiently large $$|z|, z \in Q$$ we have that $$|g(z)|/|h(z)| \leq 1$$ 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!