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Complex identity

  1. Mar 17, 2012 #1
    Hi there.
    I need help simplifying the following:
    |[itex]\sqrt{z^2 -1}[/itex] +z| + |[itex]\sqrt{z^2 -1}[/itex] -z|

    What I did was I rewrote z in polar coordinates, but I ran into some difficulties taking the square root of [itex]r^2(cos2θ-sin2θ)-1[/itex].
    I also tried rewritting z in exponential form, but also had problems. Help?
     
  2. jcsd
  3. Mar 17, 2012 #2
    In a first approach to the problem, I'd suggest you try to use the triangle inequality, both for the complete expression and for each of the terms.

    You might end up with 2 inequalities in the form: "expression" less or equal to "a + b", and "expression" greater or equal to "a + c", hence "expression" equal to "a"

    EDIT: I've tried and that doesn't work, sorry :P
     
  4. Mar 17, 2012 #3
    I think I've got it!

    If you square the whole expression and use a couple of identities on conjugates you end up with this:

    |[itex]\sqrt{z^2 -1}[/itex] +z| + |[itex]\sqrt{z^2 -1}[/itex] -z|= [itex]\sqrt{4z^2-1}[/itex]
     
  5. Mar 18, 2012 #4
    I got 4z^2-4 not 4z^2-1
     
  6. Mar 18, 2012 #5
    Actually I've just discovered a mistake in my calculation,

    The result is: 2z
     
  7. Mar 18, 2012 #6
    Oh, ok. I ended up getting 2z as well.
    But what doesnt make sense to me is that the original value should have no i (imaginary number) in it since we're taking the modulus. But after simplifying it, we get 2zm which would still have i in it.
     
  8. Mar 18, 2012 #7
    I don't see why that doesn't make sense to you.

    You can take the modulus of any complex number, and "z" is a complex number, it has an "i" in it: z=x+iy with real part x and imaginary part y.

    The modulus of a complex number is its distance to the origin defined as |z|= sqrt(x^2 + y^2) = r
     
  9. Mar 18, 2012 #8

    AlephZero

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    It doen't make sense because the value of the |...| function is a real number, but your result of 2z is a complex number.

    You can't just ignore the imaginary part of 2z, without a proper mathmatical reason for doing so.
     
  10. Mar 18, 2012 #9
    Oh you're completely right, when I said 2z I meant 2|z| of course

    But still I think the solution is not completely right! (Sorry, this is what happens when you try to solve problems quickly on a paper napkin). Let's review all the steps properly:

    [tex]A=\left |{\sqrt{z^2-1}+z}\right |+ \left |{\sqrt{z^2-1}-z}\right |[/tex]

    [tex]A^2 = \left |{\sqrt{z^2-1}+z}\right |^2 + \left |{\sqrt{z^2-1}-z}\right |^2 + 2\left |{\sqrt{z^2-1}+z}\right |\left |{\sqrt{z^2-1}-z}\right |[/tex]

    [tex]A^2 = (\sqrt{z^2-1}+z)\overline{(\sqrt{z^2-1}+z)}+ (\sqrt{z^2-1}-z)\overline{(\sqrt{z^2-1}-z)}+2[/tex]

    [tex]A^2= (\sqrt{z^2-1})(\overline{\sqrt{z^2-1}}) + \overline{z}z + \overline{z}\sqrt{z^2-1} + z\overline{\sqrt{z^2-1}}+ [/tex]

    [tex]+ (\sqrt{z^2-1})(\overline{\sqrt{z^2-1}}) + \overline{z}z - \overline{z}\sqrt{z^2-1} - z\overline{\sqrt{z^2-1}} + 2[/tex]

    [tex]A^2 = 2\left |{\sqrt{z^2-1}}\right |^2+2\left |{z}\right |^2 +2[/tex]

    [tex]A^2= 2\left |{z^2-1}\right |+2\left |{z^2}\right |+2[/tex]

    In conclusion:

    [tex]z^2 \geq{1} \rightarrow{A^2=4z^2}\Longrightarrow{A=2\left |{z}\right |}[/tex]

    [tex]z^2<1 \rightarrow{A^2=4}\Longrightarrow{A=2}[/tex]
     
    Last edited: Mar 18, 2012
  11. Mar 18, 2012 #10
    I solved this a little earlier and got the same thing. Thanks for the help!
     
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