Triangle inequality w/ Complex Numbers

In summary: Right now, you are assuming that the inequality is true and then saying it follows from the triangle inequality, which is circular reasoning.
  • #1
mynameisfunk
125
0
given z, w[tex]\in[/tex]C, and |z|=([conjugate of z]z)1/2 , prove ||z|-|w|| [tex]\leq[/tex] |z-w| [tex]\leq[/tex] |z|+|w|

I squared all three terms and ended up with :
-2|z||w| [tex]\leq[/tex] |-2zw| [tex]\leq[/tex] 2|z||w|
I know this leaves the right 2 equal to each other but i figured if i show that since there exists a z[tex]\geq[/tex]w[tex]\geq[/tex]0, then |z-w| > |z|+|w| would be impossible.

Can someone tell me if they think I screwed up or I am not done?
 
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  • #2
You assumed what you're trying to prove, so that doesn't really work.

Note that for any two complex numbers z and w, we have that zw* + z*w ≤ 2|z||w|. Now,
|z+w|2 = (z+w)*(z+w) = (z*+w*)(z+w) = zz* + zw* + z*w + ww* ≤ |z|2 + 2|z||w| + |w|2 = (|z|+|w|)2. This proves that |z+w| ≤ |z|+|w|. Can you prove the rest from here?

Edit: Fixed less than or equal to signs.
 
  • #3
Thank You for the help. I have redone the proof. here is what I've got:

It is obvious that |z|-|w|[tex]\leq[/tex]|z|, which yields |z|-|w|[tex]\leq[/tex]|z-w|. From this, we have that ||z|-|w||[tex]\leq[/tex]|z-w|. This shows the left inequality. The right-side I will show by contradiction. Assume |z-w|>|z|+|w|, this implies that w<0[tex]\Rightarrow[/tex]|z-w|=|z|+|w|>|z|+|w|, which is a contradiction showing that |z-w| is at most equal to |z|+|w|. Done.

Could anyone let me know if they think this is ok?
 
  • #4
mynameisfunk said:
It is obvious that |z|-|w|[tex]\leq[/tex]|z|, which yields |z|-|w|[tex]\leq[/tex]|z-w|.

How do you use the first inequality to arrive at the second one? I'm not following your reasoning here.

From this, we have that ||z|-|w||[tex]\leq[/tex]|z-w|.

You can do better than that. Use the fact that |z-w| ≥ |z|-|w| and |z-w| ≥ |w|-|z| to arrive at your conclusion.

The right-side I will show by contradiction. Assume |z-w|>|z|+|w|, this implies that w<0[tex]\Rightarrow[/tex]|z-w|=|z|+|w|>|z|+|w|, which is a contradiction showing that |z-w| is at most equal to |z|+|w|.

First, you need to remember that you're working with complex numbers here, so you can't say anything concerning their ordering. Second, you've assumed a lot of things which aren't necessarily true so I would scrap this approach. Everything that you want to derive for this part follows from the inequality |z+w| ≤ |z|+|w|.
 
  • #5
[tex]||z|-|w|| \leq |z-w| \leq |z|+ |w|[/tex]

pf: Left Side:
[tex]|z| \leq |z-w+w|[/tex]
[tex]|z|-|w| \leq |z-w|[/tex]
[tex]||z|-|w|| \leq |z-w|[/tex]

Right Side:
[tex]|z| \leq |z| + 2|w|[/tex]
[tex]|z-w| \leq |z| + |w|[/tex]

Hows that?
 
  • #6
I'm not sure if you are hiding the use of the actuall formula for |z| in your proof but to me, it looks like you are just assuming the result is true to prove it is true.
 
  • #7
[tex]|z| = (z\bar{z})^{\frac{1}{2}}[/tex]

Is that what you were asking for?

Also, I now see that the right side is true by the triangle inequality.
 
Last edited:
  • #8
You can't say it's true by the triangle inequality because the triangle inequality is what you are proving. You need to prove that your definition of |z| implies [tex] ||z|-|w||\leq|z-w|\leq|z|+|w|[/tex].
 

1. What is the Triangle Inequality Theorem for Complex Numbers?

The Triangle Inequality Theorem states that the absolute value of the sum of two complex numbers is always less than or equal to the sum of the absolute values of the individual complex numbers. In other words, for complex numbers a + bi and c + di, |a + bi + c + di| ≤ |a + bi| + |c + di|.

2. How is the Triangle Inequality Theorem used in complex analysis?

The Triangle Inequality Theorem is used to set bounds on the magnitude of complex numbers. This is particularly useful in proving convergence of infinite series and in solving problems involving complex numbers, such as finding the roots of polynomials.

3. Can the Triangle Inequality Theorem be extended to more than two complex numbers?

Yes, the Triangle Inequality Theorem can be extended to any number of complex numbers. It states that the absolute value of the sum of n complex numbers is always less than or equal to the sum of the absolute values of the individual complex numbers.

4. How does the Triangle Inequality Theorem relate to the concept of a triangle in geometry?

The Triangle Inequality Theorem is named after the geometric concept of a triangle because it is based on the idea that the shortest distance between two points is a straight line. In a triangle, the sum of the lengths of any two sides must be greater than the length of the third side, which is similar to the Triangle Inequality Theorem for complex numbers.

5. Are there any real-world applications of the Triangle Inequality Theorem for Complex Numbers?

Yes, the Triangle Inequality Theorem has many real-world applications, particularly in engineering and physics. It is used in signal processing, control systems, and circuit analysis to set limits on the amplitude of signals and to ensure stability in systems. It is also used in navigation and telecommunications to calculate the shortest path between two points.

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