Question

Show that for all z on the circle |z|=3
\(\dfrac{1}{120}\leq |\dfrac{1}{z^4-4z^2+3}|\leq \dfrac{1}{48}\)
Please, help

Answer 1

@ganeshie8

Answer 2

I was thinking about using triangle inequality
\[|z|^4-4|z|^2-|3|\le |z^4-4z^2+3| \le |z|^4+4|z|^2+|3| \\ 3^4-4(3)^2-3 \le |z^4-4z^2+3| \le 3^4+4(3)+3\]
then flipping things changes the direction of the inequality

Answer 3

I guess we could have done this instead on the that one side instead:
\[|a+b+c| >|a|-|b+c|>|a|- \{|b|-|c|\}=|a|-|b|+|c| \\ \text{ instead of } \\ |a+b+c|>|a+b|-|c|>|a|-|b|-|c|\]

Answer 4

since the first way gives us what we want

Answer 5

The far left is 42, not 48 as we want.

Answer 6

yep did you read my recent post

Answer 7

Yes, I do. Thank you so much. ha!! I don't think it is that simple. hehehe ...

Answer 8

why not?

Answer 9

We have a similar problem in class and my prof gave us a very complicated explanation for that. I would like to know if there is any shortcut or not. That's why I post it here.

Answer 10

but why don't you think my way works?

Answer 11

I know you heard of the triangle inequality. That is what I used above.

Answer 12

I'm confused when you said "I don't think it is that simple." did you mean to say "I didn't think it would be that simple."

Answer 13

@myininaya I didn't mean your method is wrong. I apology if I made you think that I don't believe your method. It is perfectly right. I got it.

Answer 14

alright I just did misunderstood above

Answer 15

I know how to use triangle inequality. I just don't think of it on this problem.

Answer 16

Overestimate the problems is my "disease"

Answer 17

i think we all do that