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:
10 green
Step-by-step explanation:
24 total balls. 17 Not red, so 24-17=7. 7 Red balls. That leaves 17 balls. 10 of them are neither yellow nor blue, and since the amount of red balls has been figured out, we do not need to account for them. 17 balls - 10 not yellow or blue equals 7 balls that can be yellow or blue. 17 balls left, minus the 7 yellow or blue balls equals 10.
ANSWER
EXPLANATION
The given expression is
The middle radical contains a perfect square.
Note that:
