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Homework Help: Complex inequality

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data


    [tex]|Re(z)| + |Im(z)| \le \sqrt{2}|z|[/tex]

    3. The attempt at a solution

    I was going to try to get from the LHS to RHS

    I can only see squaring then square-rooting the LHS and somehow getting to

    something involving [tex]\sqrt{|Re(z)|^2 + |Im(z)|^2}[/tex] to recover |z|, but I don't see it... any hints?
  2. jcsd
  3. Jan 10, 2010 #2
    Can you find an equality relating just |Re(z)| and |z|?
  4. Jan 10, 2010 #3

    [tex]z + \bar{z} = 2 Re(z)[/tex]

    [tex]|z +\bar{z}| = 2 |Re(z)|[/itex]

    [tex]|z| + |\bar{z}| \ge 2 |Re(z)|[/tex]

    How about that?
  5. Jan 10, 2010 #4
    re(z) = |z|cos(x)
    im(z) = |z|sin(x)

    use trig identities
  6. Jan 10, 2010 #5
    cscott, your last inequality is almost the right one, in fact, you can get the one I'm thinking of from it. You're making it a bit more complicated than it needs to be though.
  7. Jan 10, 2010 #6
    How about,

    [tex]Re(z) + Im(z) = |z| \left (\cos \theta + \sin \theta \right )[/tex]

    cos+sin has a maximum absolute value of [itex]\sqrt{2}[/itex] so,

    [tex]|Re(z)| + |Im(z)| \le \sqrt{2}|z|[/tex]

    I know I'm still making this more difficult than necessary :s
  8. Jan 10, 2010 #7


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    Gold Member

    That works. If [itex]z = a + bi[/itex] then you can also prove the inequality from the fact that [itex](a-b)^2 \geq 0[/itex].
  9. Jan 10, 2010 #8
    It definitely works. The inequality I was looking for was |Re(z)|<=|z|, by the way.
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