# Triangle inequality for complex numers

1. Oct 26, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
show
|(|z|-|z'|)|<=|z-z'|

3. The attempt at a solution

I used z=a+ib and z'=a'+ib' and ended up with the reverse inequality to the above by proving (ab'-ba')^2>=0 hence the reverse of the sign above. Where have I gone wrong?

2. Oct 26, 2007

### cristo

Staff Emeritus
Well, I don't know what you did wrong if I can't see what you did!

3. Oct 27, 2007

### pivoxa15

Good point. Bascially I expanded the complex numbers out and squared them.

4. Oct 28, 2007

### cristo

Staff Emeritus
Ok, you're not getting the point here. Show me exactly what you did, in algebra, and I can see whether you are correct or not.

5. Oct 28, 2007

### rock.freak667

Think of the numbers like vectors and draw them on an argand diagram...you'll get a triangle...and two sides added should be greater than the third side...so....vice versa

6. Oct 30, 2007

### pivoxa15

I found my mistake. I multiplied by a negative number along the way so the sign was flipped and I didn't account for that when I first did it.