Analytic Function Root Finding - Rouche's Theorem

Question

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 $|z|<1/2$.

Answer 1

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 $|z|=\frac12$.

Answer 2

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 $\,|z|\le\frac{3}{4}\,$ ...

Answer 3

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.

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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 $|z| < 1$.