# (Complex number) I have no idea on this

• Saitama

#### Saitama

(Complex number) I have no idea on this :(

## Homework Statement

If z lies on circle |z|=2, then show that
$$\left\lvert \frac{1}{z^4-4z^2+3} \right\rvert ≤ \frac{1}{3}$$

## The Attempt at a Solution

Please somebody give me an idea. [URL]http://209.85.48.12/11451/115/emo/dizz.gif[/URL]

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Consider any fraction x/y where x,y>0. If you replace y with something smaller, you make the fraction larger. In other words, if y>u, then x/y<x/u. Can you find something that's smaller than $|z^4-4z^2+3|$?

I haven't verified that that this gives us something useful, but it seems like the obvious place to start.

Can you find something that's smaller than $|z^4-4z^2+3|$?

No. How will i do that?

Sorry, I didn't notice this was for complex numbers.

If |z| <= 2, |z|^2 = |z^2| <= 4, ...

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No. How will i do that?
The triangle inequality. $$|z|-|w|\leq|z+w|\leq|z|+|w|$$

The triangle inequality. $$|z|-|w|\leq|z+w|\leq|z|+|w|$$

Okay, but how i will apply it here?

Should it be like this:-
$$|z^4-4z^2+3|≤ |z^4|+|3-4z^2|$$

Okay, but how i will apply it here?

Should it be like this:-
$$|z^4-4z^2+3|≤ |z^4|+|3-4z^2|$$
That inequality is correct, but you need to find something smaller than (or equal to) $|z^4-4z^2+3|$. I don't think I can tell you much more without solving the whole problem for you. I'll just add that I have verified that this approach gives us the correct answer.

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I think it's easier to just use the triangle inequality twice.

That inequality is correct, but you need to find something smaller than (or equal to) $|z^4-4z^2+3|$. I don't think I can tell you much more without solving the whole problem for you. I'll just add that I had verified that this approach gives us the correct answer.

$$|z^4-4z^2+3|≥ |z^4|-|3-4z^2|$$

$$|z^4-4z^2+3|≥ |z^4|-|3-4z^2|$$
Yes. Now do something similar to the second term on the right, to get something even smaller than (or equal to) $|z^4|-|3-4z^2|$.

Yes. Now do something similar to the second term on the right, to get something even smaller than (or equal to) $|z^4|-|3-4z^2|$.

How will i do that? How will i apply the triangle inequality?

Note that z4-4z2+3=(z2-1)(z2-3), and use de Moivre's formula.

ehild

I tried that but i am not able to understand how would i apply De Moivre's theorem?

What is te magnitude of z2?

ehild

Magnitude of z2? Now you are making me confused (which i am already).

Magnitude is the same as absolute value. What is it for z?

ehild

For z its |z| or $\sqrt{x^2+y^2}$
and for z2, it is $x^2+y^2$

Right..?

Hi Pranav-Arora!

Can you also write z as an exponential power?

So what can you say about |z2|?

Hi Pranav-Arora!

Can you also write z as an exponential power?

So what can you say about |z2|?

Hi ILS! :)

|z|=2
therefore |z2|=4. :)

Yep! [URL]http://i154.photobucket.com/albums/s271/R2Garnets/Smileys/Smiley-Duh.gif[/URL]

So what can you say about |z2-1|?

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How will i do that? How will i apply the triangle inequality?
If I tell you that, I will have completely solved the problem for you. Isn't the 99% I've already done enough? You have to try something yourself.

Have you tried anything at all? I only gave you two inequalities, so if the first one you tried didn't work, it has to be the other one. It looks like you're not even trying, and just want someone else to do 100% of the work for you.

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Yep! [URL]http://i154.photobucket.com/albums/s271/R2Garnets/Smileys/Smiley-Duh.gif[/URL]

So what can you say about |z2-1|?

Should it be equal to 3?

If I tell you that, I will have completely solved the problem for you. Isn't the 99% I've already done enough? You have to try something yourself.

Have you tried anything at all? I only gave you two inequalities, so if the first one you tried didn't work, it has to be the other one. It looks like you're not even trying, and just want someone else to do 100% of the work for you.

No Fredrik, i am trying myself too. But i am really not able to work out what should i do next?
Complex number is the weakest point of mine in mathematics and it is really really hard for me. I try to do my best but if you think that you have told me the 99% of the solution, i will try it again. :)

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Should it be equal to 3?

No. How did you get to 3?

No. How did you get to 3?

I thought |z2|=4.
or |z2|-1=3

Since |z2|-1=|z2-1|
therefore |z2-1|=3

EDIT: I have realized that |z2|-1=|z2-1| does not hold true at 0.

What should i do now? :(

Since |z2|-1=|z2-1|

This is not true.
Why isn't it?

Consider that z is not a real number, but a complex number.
It is represented by a 2D vector.
The absolute value is not the regular absolute value of real numbers, but the length of the vector that z represents.

This is not true.
Why isn't it?

Because it does not hold true at 0.

No Fredrik, i am trying myself too.
You haven't shown any of your work. When it seems to you that a method that we have suggested may not work, you have to post your reasons for thinking so.

But i am really not able to work out what should i do next?
I don't know how that's possible, since I gave only gave you two equalities and pointed out the specific term you should use one of them on. There are only two things you can even try. What do you get when you try those two things?

Perhaps you can apply the triangle inequality?

Consider that z is not a real number, but a complex number.
It is represented by a 2D vector.
The absolute value is not the regular absolute value of real numbers, but the length of the vector that z represents.

I know that but how would it help in finding |z2-1|?

I know that but how would it help in finding |z2-1|?

Let's take a step back and start at |z-1|.
And let's say that z=x+iy.

Geometrically z is represented by the vector (x,y) and the number 1 is represented by the vector (1,0).
Which vector is represented by (z-1)?
And what is its length?

I realized that there's (very) tiny possibility for misinterpretation of what I said on page 1, so I will clarify. What I told you there is to continue what you started, like this:
$$|z^4-4z^2+3|≥ |z^4|-|3-4z^2|\geq\text{something}$$ by applying one of the two inequalities I posted to $|3-4z^2|$.

One more thing, I think the solution I'm suggesting is simpler than the other one that's been suggested, since that one requires you to start by solving an equation to find a factorization.

Okay, i give up.

I am not able to proceed further so i think i should leave this problem. :)
I have gone through our whole discussion around three times but i am still not able to figure out what's this "something"?

Anyways, thanks for your attention to this thread. I will try it later when i have practiced more questions on Complex numbers. :)

I still can't get my head around this question. I see now the two solution methods but neither seems nice to me. I wonder what the proper way to handle these is.

I realized that there's (very) tiny possibility for misinterpretation of what I said on page 1, so I will clarify. What I told you there is to continue what you started, like this:
$$|z^4-4z^2+3|≥ |z^4|-|3-4z^2|\geq\text{something}$$ by applying one of the two inequalities I posted to $|3-4z^2|$.

One more thing, I think the solution I'm suggesting is simpler than the other one that's been suggested, since that one requires you to start by solving an equation to find a factorization.

@Fredrik: I'm afraid this does not work:

$||3|-|4z^2|| \le |3-4z^2| \le |3|+|4z^2|$

$||3|-4|z|^2| \le |3-4z^2| \le |3|+4|z|^2$

$|3-4\cdot 2^2| \le |3-4z^2| \le |3|+4\cdot 2^2$

$13 \le |3-4z^2| \le 19$

So:
$$-3 =16 - 19 \le |z^4|-|3-4z^2| \le 16 - 13 = 3$$
This includes 0, so the result is: $||z^4|-|3-4z^2|| \ge 0$.
But we need to proof that $||z^4|-|3-4z^2|| \ge 3$

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@Fredrik: I'm afraid this does not work:
Oops, you're right. Looking at it again, I see that what I get from the $$|z^4-4z^2+3|≥ |z^4|-|3-4z^2|\geq\text{something}$$ approach is $$|z^4-4z^2+3|≥-3.$$ I did this rather quickly and must have missed the minus sign.

I still can't get my head around this question. I see now the two solution methods but neither seems nice to me. I wonder what the proper way to handle these is.

IMHO ehild's method of factoring, followed by the triangle inequalities, is the nicest one.

I was leaving it to Pranav-Arora to appreciate your picture. He likes pictures!