# Triangle inequality w/ Complex Numbers

1. Sep 9, 2010

### mynameisfunk

given z, w$$\in$$C, and |z|=([conjugate of z]z)1/2 , prove ||z|-|w|| $$\leq$$ |z-w| $$\leq$$ |z|+|w|

I squared all three terms and ended up with :
-2|z||w| $$\leq$$ |-2zw| $$\leq$$ 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$$\geq$$w$$\geq$$0, then |z-w| > |z|+|w| would be impossible.

Can someone tell me if they think I screwed up or I am not done?

2. Sep 9, 2010

### jgens

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. Sep 9, 2010

### mynameisfunk

Thank You for the help. I have redone the proof. here is what I've got:

It is obvious that |z|-|w|$$\leq$$|z|, which yields |z|-|w|$$\leq$$|z-w|. From this, we have that ||z|-|w||$$\leq$$|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$$\Rightarrow$$|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. Sep 9, 2010

### jgens

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

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

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. Oct 8, 2010

### mynameisfunk

$$||z|-|w|| \leq |z-w| \leq |z|+ |w|$$

pf: Left Side:
$$|z| \leq |z-w+w|$$
$$|z|-|w| \leq |z-w|$$
$$||z|-|w|| \leq |z-w|$$

Right Side:
$$|z| \leq |z| + 2|w|$$
$$|z-w| \leq |z| + |w|$$

Hows that?

6. Oct 9, 2010

### Wizlem

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. Oct 9, 2010

### mynameisfunk

$$|z| = (z\bar{z})^{\frac{1}{2}}$$

Is that what you were asking for?

Also, I now see that the right side is true by the triangle inequality.

Last edited: Oct 9, 2010
8. Oct 10, 2010

### Wizlem

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 $$||z|-|w||\leq|z-w|\leq|z|+|w|$$.

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