Pleeeeeeeeese help quik

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

alexandr1967 [171]

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.

6 0

3 years ago

What is 4(-2)+(-3)(-5)?

Svetlanka [38]

I say that the answer is 7 bc it’s -8+15

8 0

3 years ago

Read 2 more answers

A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of

Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

the angle of elevation is Ф = 45° and
the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

the angle of elevation is Ф = 45° and
the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So, y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

5 0

2 years ago

( – 1, – 5) 7x+2y=13, 4x+4y=14

serious [3.7K]

Answer:

{x,y} = {6/5,23/10}

Step-by-step explanation:

[1] 7x + 2y = 13

[2] 4x + 4y = 14 <---------- linear equations given

Graphic Representation of the Equations : PICTURE
2y + 7x = 13 4y + 4x = 14

Solve by Substitution :

// Solve equation [2] for the variable y

[2] 4y = -4x + 14

[2] y = -x + 7/2

// Plug this in for variable y in equation [1]

[1] 7x + 2•(-x +7/2) = 13

[1] 5x = 6

// Solve equation [1] for the variable x

[1] 5x = 6

[1] x = 6/5

// By now we know this much :

x = 6/5

y = -x+7/2

// Use the x value to solve for y

y = -(6/5)+7/2 = 23/10

// Plug this in for variable y in equation [1]

[1] 7x + 2•(-x +7/2) = 13

[1] 5x = 6

// Solve equation [1] for the variable x

[1] 5x = 6

[1] x = 6/5

// By now we know this much :

x = 6/5

y = -x+7/2

// Use the x value to solve for y

y = -(6/5)+7/2 = 23/10

4 0

3 years ago

Help as soon as possible please

marta [7]

(3x + 5)° + (10x - 7)° = 180°

13x + (-2) = 180

13x = 180 + 2

13x = 182

x = 14

∠QRT = 3 * 14 + 5 = 42 + 5 = 47°

∠TRS = 10 * 14 - 7 = 140 - 7 = 133°

6 0

4 years ago