Rouché for Polynomial $p(z)=z^7+z(z-3)^3+1$ around non-centered annulus.

Question

Let $p(z)=z^7+z(z-3)^3+1$. Find the numbers of zeros (including multiplicities) in $B_{1}(3)$.

I want to apply Rouché's Theorem for $p(z)$ and $g(z)=-z(z-3)^3$, but I don't know how to confine on $|z-3|=1$. The problem should be solved without using a calculator.

Advice appreciated. Thanks

Answer 1

It's better to work with unit circle $|z|\leqslant1$, For this purpose let $w=z-3$ then $p(z)=(w+3)^7+w^4+3w^3+1$ and for $|w|=1$ we have $$|f(w)|=|w^4+3w^3+1|\leqslant|w|^4+3|w|^3+1=5<|w+3|^7=|g(w)|$$ therefore the number of zeros $p(w)$ and $g(w)$ in $|w|\leqslant1$ are the same, but $g(w)$ has no zeros there.

Answer 2

You get $$ |z^7-1|\ge 2^7-1=127 $$ and $$ |z(z-3)^3|\le 4 $$ on the circle $|z-3|=1$. Thus $p$ has the same number of roots inside the circle as $z^7+1$.

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Roots of $z^7+\alpha z(z-3)^3+1$ for $α\in[0,1]$, red for $α=1$.

Answer 3

Let $q(z) = z^7.$ Then on the circle $z= 3 +e^{it}, 0\le t \le 2\pi,$

$$|p(z)-q(z)| = |z(z-3)^3 +1| \le |z(z-3)^3| +1 \le 4\cdot 1^3 + 1 = 5 .$$

But note $|q|\ge 2^7$ on this circle. Since $5< 2^7,$ $p,q,$ have the same number of zeros in $D(3,1).$ Since $q=0$ has no zeroes in this disc, $p$ has no zeros in this disc by Rouche.