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Proving on modulus and conjugates

  1. Nov 15, 2005 #1
    I am proving that (sqrt)2 |z| >= |Re z| + |Im z|. The professor gave this as an example, but I want to ask something...

    Why is it that I can't use this strategy:
    -> |Re z| <= |z|
    -> |Im z| <= |z|
    -> adding corresponding expression yields |Re z| + |Im z| <= 2|z|. (what's wrong here?)
  2. jcsd
  3. Nov 15, 2005 #2

    Tom Mattson

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    You can't use it because [itex]|Re z|+|Im z| \leq 2|z|[/itex] simply does not imply that [itex]|Re z| + |Im z| \leq \sqrt{2}|z|[/itex]. Simple example: Say [itex]|Re z|+|Im z|=1.5[/itex]. The first inequality is satisfied, but the second is not.
  4. Nov 15, 2005 #3


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    Try squaring both sides.

    That is an upper bound for |Re z| + |Im z|, but not the least upper bound. Try squaring both sides.
  5. Nov 15, 2005 #4


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    Oh yeah, put z=x+iy first.
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