Here is my question:
State Rouché's theorem. How many roots of the polynomial $p(z) = z^8 + 3 z^7 + 6 z^2 + 1$ are contained in the annulus {$1 < |z| < 2$}?
The statement is fine.
I then consider the functions $g_1 (z) = z^8 + 3z^7 + 1$ and $f_1(z) = 6z^2$, giving one root of $p = f_1 + g_1$ inside {$|z| < 1$}, as $f_1(z) = 0 \iff z=0$.
I then consider the functions $g_2(z) = z^8 + 6z^2 + 1$ and $f_2(z) = 3z^7$, giving one root of $p = f_2 + g_2$ inside {$|z| < 2$}, as $f_2(z) = 0 \iff z=0$.
Thus I have no roots inside {$1<|z|<2$}.
Is this correct? It's a past exam question, so I'm surprised that there would be no roots inside the given annulus…
Thanks in advance! 🙂
Best Answer
I think you're forgetting about multiplicity. For example, in $B(0, 1)$, $6z^2$ has two zeroes (counted with multiplicity), so $p$ has two zeroes (counted with multiplicity) in $B(0, 1)$.
Also, once you've determined how many zeroes $p$ has in $B(0, 1)$ and $B(0, 2)$, then you know how many zeroes it has in $B(0, 2)\setminus B(0, 1) = \{z \in \mathbb{C} \mid 1 \leq |z| < 2\}$. You should then explain why none of those zeroes lie on $S^1$.