Analytic Function Root Finding - Rouche's Theorem Please help determine the number of roots of 
$$
z^7+2z^3+1
$$
in the region $1/2\leq|z|<1$.  
It seems like everything I do with Rouche's theorem does not give a strict inequality for when $|z|=1$.  I think I found that there are no roots in the disk $|z|<1/2$.  
 A: You're right: there are no roots in $\,|z|\le \frac{1}{2}\,$ , and for the other part I propose the following idea you shall develop:
For $\,|z|\le \frac{3}{4}\,$ define $\,f(z):=1+2z^3\;,\;\;g(z):=z^7\,$ , so for$\,|z|=\frac{3}{4}\,$ we get
$$|f(z)|\ge 1- 2\cdot\left(\frac{3}{4}\right)^3>\left(\frac{3}{4}\right)^7=|g(z)|$$
So $\,f+g\,$ has the same number of roots as $\,f\,$ in $\,|z|\le\frac{3}{4}\,$ , and since
$$z^3=-\frac{1}{2}\implies |z|=\sqrt[3]{\frac{1}{2}}>\frac{3}{4}$$
we already know there are no roots in $\,|z|\le\frac{3}{4}\,$ ...
A: Rouché's theorem is still sometimes valid even when the inequality is not strict.

Rouché's Theorem.
Suppose $P(z)$ and $Q(z)$ are analytic interior to a simple closed Jordan curve $C$ and continuous on $C$.  If $$F(z) = P(z) + Q(z) \neq 0$$ and $$|P(z)| \leq |Q(z)|$$ on $C$, then $F(z)$ has the same number of zeros interior to $C$ as does $Q(z)$.

Most proofs of Rouché's theorem (or at least the two that I'm thinking of) can be made to render this result after a minor modification.  Essentially the strict inequality $|P| < |Q|$ is used to guarantee that $F$ is nonzero on $C$, so by simply assuming this condition we can relax the inequality.
This particular statement of Rouché's theorem is a synthesis of Theorem 1.3 and Exercise 7 of Section 1 in Marden's Geometry of Polynomials.

The problem at hand.
By replacing $z$ by $e^{i\theta}$ we calculate
$$
\left|2z^3 + 1\right|^2 = 4\cos(3\theta) + 5.
$$
Now $$4\cos(3\theta) + 5 \geq 1$$ and $$4\cos(3\theta) + 5 = 1$$ when $\theta = \pm \frac{\pi}{3},\pi$, so on $|z| = 1$ we conclude that
$$
\left|2z^3 + 1\right| \geq 1 = \left|z^7\right|
$$
and
$$
\left|2z^3 + 1\right| = 1 \quad \Longleftrightarrow \quad z = -1,\,e^{\pm i \pi/3}.
$$
We need to check that these points are not zeros of $z^7 + 2z^3 + 1$.  Indeed,
$$
\left[z^7 + 2z^3 + 1\right]_{z=-1} = -1 - 2 + 1 \neq 0
$$
and
$$
\left[z^7 + 2z^3 + 1\right]_{z=e^{\pm i \pi/3}} = e^{\pm i \pi/3} - 2 + 1 \neq 0.
$$
We may now apply Rouché's theorem, which tells us that $z^7 + 2z^3 + 1$ has the same number of zeros in $|z| < 1$ as does $2z^3 + 1$.  The polynomial $2z^3 + 1$ has three zeros on $|z| = 2^{-1/3} < 1$, so we conclude that $z^7 + 2z^3 + 1$ also has three zeros in $|z| < 1$.
A: You are correct, there are not roots in the disk $|z|<\frac12$. To make it more simple, break it down into pieces. First find, $|f(z)|\le |g(z)|$ for $|z|=1$, then repeat the same process when $|z|=\frac12$. 
