Let 𝔻 = { 𝑧 ∈ ℂ ∶ |𝑧| < 1} and 𝑓: 𝔻 → ℂ be defined by 

\(f(z)=\rm z-25z^3+\frac{z^5}{5!}-\frac{z^7}{7!}+\frac{z^9}{9!}-\frac{z^{11}}{11!}\)

Consider the following statements:

𝑃: 𝑓 has three zeros (counting multiplicity) in 𝔻.

𝑄: 𝑓 has one zero in 𝕌 = { 𝑧 ∈ ℂ ∶ \(\frac{1}{2}\) < |𝑧| < 1}.

Then 

Concept:

Rouche's theorem: If f(z) and g(z) are two analytic functions within and on a simple closed curve C such that |f(z)| > |g(z)| at each point on C, then both f(z) and f(z) + g(z) have the same number of zeros inside C.

Explanation:

For P,

𝔻 = { 𝑧 ∈ ℂ ∶ |𝑧| < 1} and 𝑓: 𝔻 → ℂ be defined by 

\(f(z)=\rm z-25z^3+\frac{z^5}{5!}-\frac{z^7}{7!}+\frac{z^9}{9!}-\frac{z^{11}}{11!}\)

Let g(z) = z - 25z³ and h(z) = \(\frac{z^5}{5!}-\frac{z^7}{7!}+\frac{z^9}{9!}-\frac{z^{11}}{11!}\)

then in 𝔻

|g(z)| = 1 + 25 = 26 and |h(z)| = \(\frac{1}{5!}+\frac{1}{7!}+\frac{1}{9!}+\frac{1}{11!}\)

Hence  |g(z)| > |h(z)|

Therefore by Rouche's theorem, g(z) and g(z) + h(z) have the same number of zeros inside 𝔻

Now, g(z) = z - 25z3 is a polynomial of degree 3 so g(z) has 3 zeros in 𝔻

Hence 𝑓 has three zeros (counting multiplicity) in 𝔻

P is true

For Q,

𝕌 = { 𝑧 ∈ ℂ ∶ \(\frac{1}{2}\) < |𝑧| < 1}

Let's check for |𝑧| < \(\frac{1}{2}\) 

Let g(z) = z - 25z3 and h(z) = \(\frac{z^5}{5!}-\frac{z^7}{7!}+\frac{z^9}{9!}-\frac{z^{11}}{11!}\)

then inside |𝑧| < \(\frac{1}{2}\)

|g(z)| = \(\frac{1}{2}\) + 25 × \(\frac{1}{2}\) = 13 and |h(z)| = \(\frac{(\frac12)^5}{5!}+\frac{(\frac12)^7}{7!}+\frac{(\frac12)^9}{9!}+\frac{(\frac12)^{11}}{11!}\)

Hence  |g(z)| > |h(z)|

Therefore by Rouche's theorem, g(z) and g(z) + h(z) have the same number of zeros inside |𝑧| < \(\frac{1}{2}\)

Now, g(z) = z - 25z3 is a polynomial of degree 3 so g(z) has 3 zeros in |𝑧| < \(\frac{1}{2}\)

So, 𝑓 has three zeros (counting multiplicity) in |𝑧| < \(\frac{1}{2}\) 

Therefore 𝑓 has 3 - 3 = 0 zeros in 𝕌

Q is false

𝑃 is TRUE but 𝑄 is FALSE

(1) is correct