Triangle inequality for complex numers

[pivoxa15]

Homework Statement

show
|(|z|-|z'|)|<=|z-z'|

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?

 

[cristo]

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

 

[pivoxa15]

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

 

[cristo]

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.

 

[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

 

[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.