Answered

IDNLearn.com: Your go-to resource for finding precise and accurate answers. Join our knowledgeable community and get detailed, reliable answers to all your questions.

Use the theorem in Sec. 28 to show that if f (z) is analytic and not constant throughout a domain D, then it cannot be constant throughout any neighborhood lying in D. Suggestion: Suppose that f(z) does have a constant value ωo throughout some neighborhood in D.

Sagot :

Danialaminidn

Answer:

The value of f(z) is not constant in any neighbourhood of D. The proof is as explained in the explaination.

Step-by-step explanation:

Given

For any given function f(z), it is analytic and not constant throughout a domain D

To Prove

The function f(z) is non-constant constant in the neighbourhood lying in D.

Proof

1-Assume that the value of f(z)  is analytic and has a constant throughout some neighbourhood in D which is ω₀

2-Now consider another function F₁(z) where

F₁(z)=f(z)-ω₀

3-As f(z) is analytic throughout D and F₁(z) is a difference of an analytic function and a constant so it is also an analytic function.

4-Assume that the value of F₁(z) is 0 throughout the domain D thus F₁(z)≡0 in domain D.

5-Replacing value of F₁(z) in the above gives:

F₁(z)≡0 in domain D

f(z)-ω₀≡0 in domain D

f(z)≡0+ω₀ in domain D

f(z)≡ω₀ in domain D

So this indicates that the value of f(z) for all values in domain D is a constant  ω₀.

This contradicts with the initial given statement, where the value of f(z) is not constant thus the assumption is wrong and the value of f(z) is not constant in any neighbourhood of D.

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is dedicated to providing accurate answers. Thank you for visiting, and see you next time for more solutions.