Help Solving Math Problem: Numerator < 5, Denominator > 3

[asi123]

Homework Statement

Hey guys.
I got this question and I don't know where to start.
There is a little hint, which is to show that the numerator is smaller then 5 and the enumerator is bigger then 3(1-abs(z)).
Any idea guys?
10x.

Homework Equations

The Attempt at a Solution

 

[Mark44]

What exactly is the question? Are you supposed to prove that
|\frac{3 + z^{3n} - |z|}{3 - z^{5n} + x^{4n} -z}| \leq \frac{5}{3(1 - |z|)}?
If so, that's not much of a hint, because it's essentially what you need to do.
If that's not what you need to do, then I have no clue.

 

[asi123]

What exactly is the question? Are you supposed to prove that
|\frac{3 + z^{3n} - |z|}{3 - z^{5n} + x^{4n} -z}| \leq \frac{5}{3(1 - |z|)}?
If so, that's not much of a hint, because it's essentially what you need to do.
If that's not what you need to do, then I have no clue.

Yeah, that's the question.

Well, how do you do that?

 

[Mark44]

I don't know how to prove this. I've spent a little time, but nothing seems to pop out at me as an approach. Before I spend any more time on it, is there any more to this problem? All I know so far is that z is complex, |z| <= 1, and that you are to prove the inequality above, which you neglected to mention in your first post.

Is there anything else you have neglected to mention?

 

[HallsofIvy]

A major problem is that is it NOT true! If z= 1, the denominator on the right side is 0 while the denominator on the left is not. If z is close to 1, then the right side will be extremely large while the left side is close to 3/2.

 

[asi123]

I don't know how to prove this. I've spent a little time, but nothing seems to pop out at me as an approach. Before I spend any more time on it, is there any more to this problem? All I know so far is that z is complex, |z| <= 1, and that you are to prove the inequality above, which you neglected to mention in your first post.

Is there anything else you have neglected to mention?

Hey.

I didn't posted the entire problem because it's in Hebrew .
I'm sorry if I didn't make my self clear. Yeah, I need to prove this inequality.
I didn't neglect any thing else.

10x.

 

[gabbagabbahey]

A major problem is that is it NOT true! If z= 1, the denominator on the right side is 0 while the denominator on the left is not. If z is close to 1, then the right side will be extremely large while the left side is close to 3/2.

I don't think that's really a problem because of the direction of the inequality... surely \frac{3}{2} \leq \infty ?

 

[gabbagabbahey]

Hey.

I didn't posted the entire problem because it's in Hebrew .
I'm sorry if I didn't make my self clear. Yeah, I need to prove this inequality.
I didn't neglect any thing else.

10x.

You did neglect something...are there any restrictions on n? Surely it has to be a positive integer?

 

[asi123]

You did neglect something...are there any restrictions on n? Surely it has to be a positive integer?

This is the entire problem (in Hebrew), I don't see any restrictions on n.
But let's say it's positive, any ideas?

10x.

 

[Citan Uzuki]

The things you need to know:

1: Since |z|≤1, we have for any positive integer k that |z|^(kn) ≤ |z| (assuming that n is positive)
2: Use the triangle inequality -- |a+b| ≤ |a| + |b| -- and its counterpart, |a-b| ≥ ||a| - |b||
3: As the problem hint suggests, prove the absolute value of the numerator is at most 5 and that the absolute value of the denominator is at least 3 - 3|z|.

 

[asi123]

The things you need to know:

1: Since |z|≤1, we have for any positive integer k that |z|^(kn) ≤ |z| (assuming that n is positive)
2: Use the triangle inequality -- |a+b| ≤ |a| + |b| -- and its counterpart, |a-b| ≥ ||a| - |b||
3: As the problem hint suggests, prove the absolute value of the numerator is at most 5 and that the absolute value of the denominator is at least 3 - 3|z|.

Ok, but in the problem, it's not |z|^(3n), it's z^(3n).
Lets talk about the numerator, I know that 3 > 3-|z| >= 2 so what I need to prove is that z^(3n) <= 2, right?
Well, how can I prove just a thing?

10x.

 

[Citan Uzuki]

Ok, but in the problem, it's not |z|^(3n), it's z^(3n).

|z^(3n)| = |z|^(3n)

Lets talk about the numerator, I know that 3 > 3-|z| >= 2 so what I need to prove is that z^(3n) <= 2, right?
Well, how can I prove just a thing?

10x.

No, you're trying to provide an upper bound on the magnitude of the numerator, so proving 3 ≥ 3-|z| ≥ 2 is not helpful. Also, you do not need to prove that z^(3n) ≤ 2, and indeed, since z is a complex number, it may not be comparable to 2 (a≤b only makes sense if a and b are both real).

What you need to prove is: |3 + z^(3n) - |z|| ≤ 5

The first step here is to apply the triangle inequality, so we have:

|3 + z^(3n) - |z|| ≤ |3| + |z^(3n)| + |-|z|| = 3 + |z|^(3n) + |z|.

so it suffices to show that 3 + |z|^(3n) + |z| ≤ 5. Can you see how to do that with what I've showed you?

Obvious hint:

Spoiler

Note that |z|^(3n) ≤ |z| ≤ 1

 

[asi123]

|z^(3n)| = |z|^(3n)



No, you're trying to provide an upper bound on the magnitude of the numerator, so proving 3 ≥ 3-|z| ≥ 2 is not helpful. Also, you do not need to prove that z^(3n) ≤ 2, and indeed, since z is a complex number, it may not be comparable to 2 (a≤b only makes sense if a and b are both real).

What you need to prove is: |3 + z^(3n) - |z|| ≤ 5

The first step here is to apply the triangle inequality, so we have:

|3 + z^(3n) - |z|| ≤ |3| + |z^(3n)| + |-|z|| = 3 + |z|^(3n) + |z|.

so it suffices to show that 3 + |z|^(3n) + |z| ≤ 5. Can you see how to do that with what I've showed you?

Obvious hint:

Spoiler

Note that |z|^(3n) ≤ |z| ≤ 1

Yeah, I can see that, 10x a lot.