Fall 2019 - Problem 9

Solution.

On $\abs{z} = 1$, write $P\p{z} = 10z + \p{z^6 - 6z^2 + 2}$, and so

In particular, $P$ does not vanish when $\abs{z} = 1$, so by Rouché's theorem, $10z$ and $P\p{z}$ have the same number of zeroes in $\D$, namely $1$. Similarly, for $\abs{z} = 2$, write $P\p{z} = z^6 + \p{-6z^2 + 10z + 2}$. Observe that

so $P$ does not vanish on $\abs{z} = 2$, and by Rouché's theorem again, $P\p{z}$ has $6$ roots in $\set{\abs{z} < 2}$. Thus, $P\p{z}$ has $6 - 1 = 5$ roots in the annulus.