Number of zeros of $p(z)=z^5+10z-3$ on annulus $A(0,1,2)$

Question

How many zeros does the function $p(z)=z^5+10z-3$ have on the annulus $A(0,1,2)$?

For $z\in C(0,2)$, we have $|f(z)|=|z^5|=32$ and $|g(z)|=|10z-3|\leqslant 10\cdot 2+3=23$, thus $|f|<|g|$ on $C(0,2)$, which implies by Rouché's Theorem that $p=f+g$ has as much zeros as $f$ in $B(0,2)$, which is $5$.

We can do the same trick on $C(0,1)$ using $f(z)=10z$ and $g(z)=z^5-3$, to show that $p$ has $1$ zero in $B(0,1)$. (Since $p(0)p(1/2)<0$, this is a real zero)

Now I want to conclude that in the annulus $A(0,1,2)$ there are $5-1=4$ zeros. However, how can I be sure that there is no zero on the circle $C(0,1)$?

Answer 1

With $f(z) = 10z$ and $g(z) = z^5-3$ you have $$ |g(z)| \le 4 < 10 = |f(z)| $$ on $C(0, 1)$. This shows not only that $p = f+g$ and $f$ have the same number of zeros in $B(0, 1)$ but also that $p$ has no zero on $C(0, 1)$: $p(z)=0$ would imply $|g(z)| = |f(z)|$.

This works generally with Rouché's theorem: If $|g(z)| < |f(z)| $ on $\partial K$ then $f+g$ and $f$ have the same number of zeros in $K$, and $f+g$ has no zeros on $\partial K$.

Answer 2

Because if $|z|=1$, then $|z^5+10z-3|\ge |10z| - |z^5-3| \ge 10 - 4 = 6$.

Answer 3

Hint: Apply the method not to $C(0,1),$ but to $C(0,1+\epsilon)$ for some small $\epsilon >0$.

Answer 4

For modulus $1$, you have $10z$ as the dominating function which has $1$ zero, and for modulus $2$, the dominating function is $z^5$ which has $5$ zeros of course thus you have a total of $5-1=4$ zeros on your given annulus.