(Complex number) I have no idea on this

[Fredrik]

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

Yes, I agree. $$|z^4-4z^2+3|=|z^2-3|\,|z^2-1|\geq \text{something}\cdot\text{something}$$ is definitely the way to go.

Pranav-Arora, the method I suggested doesn't work. We get $|z^4-4z^2+3|\geq -3$ if we do it exactly the way I suggested. We need +3 on the right, not -3, so this result is useless. Even if we change the first step into $|z^4-4z^2+3|\geq 3-|z^4-4z^2|$, we're getting something useless. I apologize for misleading you.

The only solution I have found that is similar to my original idea is to start with $$|z^4-4z^2+3|\geq|z^4-4z^2+4|-1=|z^2-2|^2-1,$$ but as you can see, the only point of doing it this way would be to make the factorization a bit easier. I don't see a way to avoid doing a factorization.

 

[verty]

It didn't help that I made a mistake in thinking that |z^4 - 4z^2| <= |z^4| - |4z^2|. It is <= |z^4| + |4z^2|. Which means that |z^4 - 4z2 + 3| <= 35, a very strange looking result. But in this question we need >=, and that I don't know how to do nicely (unless factorizing is nice but it seems lucky). I'll look at this more.

 

[I like Serena]

It didn't help that I made a mistake in thinking that |z^4 - 4z^2| <= |z^4| - |4z^2|. It is <= |z^4| + |4z^2|. Which means that |z^4 - 4z2 + 3| <= 35, a very strange looking result. But in this question we need >=, and that I don't know how to do nicely (unless factorizing is nice but it seems lucky). I'll look at this more.

If ABC is a triangle then there are 2 triangle inequalities, which combined into one are:
$$\left|~|AC| - |BC|~\right| \le |AB| \le |AC| + |BC|$$

In words:
The length of AB is greater than the difference in length of the other 2 sides.
The length of AB is less then the sum of the lengths of the other 2 sides.

You need the first form.

 

[Saitama]

I am fed up of this problem.
Now please stop posting hints and stop making me more confused.

 

[verty]

Ok, I've got my head around it now. Right.

Pranav, in case you return to this question later, here is a hint how to do it without factorizing.

If we have |a| and |b|, we can say two things: one is |a| + |b| >= |a+b|, one is ||a| - |b|| <= |a-b|. Since we need to show (what?) about |z^4 + 3 - 4z^2|, we can (do what?) using (which one?) to find the answer. :)

 

[Saitama]

I already said:-

STOP POSTING HINTS NOW.

 

[Fredrik]

Ok, I've got my head around it now. Right.

Pranav, in case you return to this question later, here is a hint how to do it without factorizing.

If we have |a| and |b|, we can say two things: one is |a| + |b| >= |a+b|, one is ||a| - |b|| <= |a-b|. Since we need to show (what?) about |z^4 + 3 - 4z^2|, we can (do what?) using (which one?) to find the answer. :)

This seems to be the exact same approach that I tried first. See my comments in #38 and #42. (Did you miss the same minus sign as I did?)

 

[verty]

It wasn't the same sign error. I didn't trust myself to apply any other form of triangle inequality, so I stuck to the traditional "hypotenuse is no greater" form. But that gave the wrong bound, and then I had a sign error trying to adapt that approach to the problem. Anyway, let's move on.

 

[Fredrik]

I didn't trust myself to apply any other form of triangle inequality, so I stuck to the traditional "hypotenuse is no greater" form.

The one with a minus sign follows from that one. For all z and w, |z+w|≤|z|+|w|. This implies that for all z and w, |z|=|(z+w)-w|≤|z+w|+|-w|=|z+w|+|w|.

So for all z and w, we have $$|z|-|w|\leq|z+w|\leq |z|+|w|.$$