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Should be a quick one

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data

    How is arg(z1/z2) = arg(z1z2) ???Where the bold z2 represents the conjugate.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 4, 2013 #2

    SammyS

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    What have you tried?

    Where are you stuck ?
     
  4. Aug 4, 2013 #3
    Yeah I tried. I just took an example say z1= 2+2i and z2= 3+4i

    So arg (z1/z2) = .56-.08i
    now arg(z1z2) = 14-2i
    I don't get how this person wrote that arg(z1/z2) = arg(z1z2)
     
  5. Aug 4, 2013 #4

    SammyS

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    Actually:

    arg (z1/z2) = arg(.56-.08i)

    and

    arg(z1z2) = arg(14-2i) .

    Are not those the same; arg(.56-.08i) and arg(14-2i) ?
     
  6. Aug 4, 2013 #5
    Opps. Thanks
     
  7. Aug 4, 2013 #6
    So you are happy that your particular example satisfies the identity [itex]arg(z_1/z_2)=arg(z_1 \bar{z_2})[/itex], but do you understand why it is an identity - ie it is satisfied by any [itex]z_1[/itex] and [itex]z_2[/itex] (with [itex]z_1,z_2 \neq 0[/itex])?

    If not, think about how you can represent each of the following:
    [tex]arg(z_1/z_2)[/tex][tex]arg(z_1 z_2)[/tex][tex]arg(\bar{z_2})[/tex] in terms of [itex]arg(z_1)[/itex] and [itex]arg(z_2)[/itex].
     
    Last edited: Aug 4, 2013
  8. Aug 4, 2013 #7

    LCKurtz

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    Of course, that just verifies it for those two particular values in that example. What happens if you write for general ##z_1## and ##z_2## in polar form and try it?
     
  9. Aug 4, 2013 #8

    [tex]arg(z_1/z_2)= arg(z1)-arg(z2)[/tex]
    [tex]arg(z_1 z_2)= arg(z1) +arg(z2)[/tex]
    [tex]arg(\bar{z_2})= -arg(z2)[/tex]but you said in terms of ##arg(z1) ##and ## arg(z2)##
    So I don;t know about the last one.

    What do you mean?
     
  10. Aug 4, 2013 #9

    LCKurtz

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    ##z = re^{i\theta}## form.
     
  11. Aug 4, 2013 #10
    That's right, you've answered all three correctly.

    So,
    [tex]arg(z_1/z_2)=arg(z_1)-arg(z_2)[/tex][tex]=arg(z_1)+arg(\bar{z_2})[/tex]
    [tex]=arg(z_1 \bar{z_2})[/tex]
     
  12. Aug 4, 2013 #11

    LCKurtz

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    But didn't he say he didn't understand the third one?
     
  13. Aug 4, 2013 #12
    I took that to mean that he/she was unsure about it simply because I'd said "in terms of [itex]arg(z_1)[/itex] and [itex]arg(z_2)[/itex]" and only one of these terms was necessary. A problem of the wording really, rather than not understanding the answer, IMO.

    EDIT: @Jbreezy When asked to represent an expression "in terms of A, B and C" (for example), you are not obliged to use all of A, B and C in your answer.
     
    Last edited: Aug 4, 2013
  14. Aug 4, 2013 #13
    I don't know this form. You represent a complex number like a + ib like this? what?
     
  15. Aug 4, 2013 #14
    Any complex number can be represented this way.

    [itex]r[/itex] represents the modulus (ie [itex]\sqrt{a^2+b^2}[/itex]) and [itex]\theta[/itex] represents the argument.
     
  16. Aug 5, 2013 #15

    Mentallic

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    It's equivalent to
    [tex]z=r \left(\cos(\theta)+i\sin(\theta)\right)=re^{i\theta}[/tex]
     
  17. Aug 5, 2013 #16
    Yeah I will answer this question when I get to that part in my text.Thx
     
  18. Aug 5, 2013 #17

    SammyS

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    Multiply z1/z2 by z2/z2 .

    This gives (z1z2)/(z2z2) .

    However, (z2z2) is a purely real number, so dividing by (z2z2) has no effect on the argument.
     
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