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Triangle inequality for complex numers

  1. Oct 26, 2007 #1
    1. The problem statement, all variables and given/known data

    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. jcsd
  3. Oct 26, 2007 #2


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    Well, I don't know what you did wrong if I can't see what you did!
  4. Oct 27, 2007 #3
    Good point. Bascially I expanded the complex numbers out and squared them.
  5. Oct 28, 2007 #4


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    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.
  6. Oct 28, 2007 #5


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    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
  7. Oct 30, 2007 #6
    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.
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