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:
16 and 3/4 is 7/4
Step-by-step explanation:
The answer is B) 43<span>°
Angle ADB is an inscribed angle which means arc AB has an angle twice that of angle ADB. The angle of the arc would be the same as that of the central angle AOB. So, mAOB = 86</span>°. And since, mAOB = mBOC, then mBOC = 86° and arc BC has a measure of 86° as well. Angle BDC intercepts the arc BC which means half of the angle of arc BC is mBDC. So, mBDC = 43<span>°.</span>
Answer in screen shot below
