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.
Answer:
96 units^2
Step-by-step explanation:
The formula for the area of a triangle: b*h/2
b= 12
h= 16
Work:
12*16/2
192/2
96 uints^2
What are you asking me ? Do you have a picture I can look at ?
Answer:
1 2/15
Explanation:
8 2/5 - 7 4/15 = ?
First convert the mixed numbers into improper fractions:
42/5 - 109/15 = ?
Answer:
c. We are 95% confident that, among the population of students that play basketball recreationally, the proportion who live on campus is between 0.12 and 0.18.
Step-by-step explanation:
According to the question, it is mentioned that there is 95% confidence interval and it is calculated (0.12,0.18)
So the correct interpretation is that they are 95% confident that the students who play basketballs the proportion should lies between 0.12 and 0.12
So, the option C is correct
The rest of the options are wrong
