Discussion
Let \(\mathbb{D} = \{z \in \mathbb{C} : |z| \lt 1\}\) and \(f : \mathbb{D} \to \mathbb{C}\) be defined by
\[ \begin{aligned} f(z) = z &- 25z^{3} + \frac{z^{5}}{5!} - \frac{z^{7}}{7!} \\ &+ \frac{z^{9}}{9!} - \frac{z^{11}}{11!} \end{aligned} \]
Consider the following statements:
P: f has three zeros (counting multiplicity) in \(\mathbb{D}\).
Q: f has one zero in \(\mathbb{U} = \{z \in \mathbb{C} : \frac{1}{2} \lt |z| \lt 1\}\).
Then
Answer: A
P is TRUE but Q is FALSE.
Use Rouche's theorem with the dominant term \(g(z) = -25z^{3}\). On \(|z| = 1\), \(|g| = 25\) and
\[ \begin{aligned} |f - g| &\le 1 + \frac{1}{5!} + \frac{1}{7!} + \frac{1}{9!} + \frac{1}{11!} \\ &\lt 2 \end{aligned} \]
so f has the same number of zeros as \(z^{3}\) in the unit disc: three. P is true.
On \(|z| = \tfrac{1}{2}\), \(|g| = 25/8 = 3.125\) while the remaining terms add up to less than 0.51, so f also has three zeros in \(|z| \lt \tfrac{1}{2}\). Hence all three zeros lie inside the smaller disc and none in the annulus \(\tfrac{1}{2} \lt |z| \lt 1\): Q is false. (Indeed \(f(z) \approx z(1 - 25z^{2})\) has zeros near 0 and \(\pm\tfrac{1}{5}\).)
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