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Complex number proof about z (single or double overlined/conjugate) = z.

  1. Jan 30, 2013 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem along with its solution are attached as TheProblemAndSolution.jpg. This post's focus is on part (iii), specifically.

    2. Relevant equations
    z (overlined) = z. (according to book)
    z (double overlined) = z. (according to me)

    3. The attempt at a solution
    I understand the first two parts but, is there a typo in part (iii)? I looked at Wikipedia and I believe that part (iii) intends to say double conjugate of z = z rather than single conjugate of z = z. Am I right in believing that the book I'm using is wrong?

    I would really appreciate a confirmation or denial because, I want to make sure that I am not missing something important if I just assume that the book is wrong!
     

    Attached Files:

    Last edited: Jan 30, 2013
  2. jcsd
  3. Jan 30, 2013 #2

    Dick

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    [itex]\bar{z}=z[/itex] only if z is real. [itex]\bar{\bar{z}}=z[/itex] for any z.
     
  4. Jan 30, 2013 #3

    jedishrfu

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    I can't see your jpg TheProblemAndSolution.jpg

    Can you provide a link for it or upload it?
     
  5. Jan 30, 2013 #4

    s3a

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    jedishrfu:
    Sorry, I forgot to attach the picture.

    Dick:
    z = a + bi (which is complex and not solely real) in the problem.
     
  6. Jan 30, 2013 #5

    jedishrfu

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    Yeah, it looks like the book is wrong for iii.

    z=a + bi so zbar = a - bi zbarbar = a + bi hence zbarbar = z
     
  7. Jan 30, 2013 #6

    Dick

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    By z is real, I just mean b=0 in z=a+bi.
     
  8. Jan 30, 2013 #7

    s3a

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    Sorry for the triple post.

    Look two posts down.
     
  9. Jan 30, 2013 #8

    s3a

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    Sorry for the triple post.

    Look one post down.
     
  10. Jan 30, 2013 #9

    s3a

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    Okay so, what the book most likely wanted to ask was: "Prove that z = z (double overlined)."?

    [Technically, I believe the term "bar" is incorrect since, looking at h-bar (Planck's constant divided by (2*pi)), I believe that the bar is always through the letter and not above it.]

    Edit:
    Dick:
    Yes, I know. :)
     
  11. Jan 30, 2013 #10

    Dick

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    Yes, I think the book wants you to prove [itex]\bar{\bar{z}}=z[/itex]. It's just a typo. And lots of people say 'bar' when they see the symbol. The pronunciation is not that formalized.
     
  12. Jan 30, 2013 #11

    jedishrfu

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    Yes, it was named in honor of Bar Rafaeli:

    http://en.wikipedia.org/wiki/Bar_Refaeli
     
  13. Jan 30, 2013 #12

    Dick

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  14. Jan 30, 2013 #13

    s3a

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    Wait, is that a joke or something?

    I ask because that woman is pretty young and I'm not sure but, wasn't that notation available to physicists that lived before she was born?

    Edit:
    Also, about the main focus of this thread.: Thank you both. :)
     
  15. Jan 30, 2013 #14

    Dick

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    Sure, joke. You're welcome!
     
  16. Jan 30, 2013 #15

    CompuChip

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    I was staring at the proof in your attachment, maybe I'm just dumb at the moment but it really looks like an (incomprehensible) proof for [itex]\overline{z} = z[/itex] ... it's not like the just made a typo with the bars.
     
  17. Jan 30, 2013 #16

    Dick

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    You're right. I didn't even look at the 'proof'. Pretty incoherent. I'm sure s3a will do better.
     
  18. Jan 30, 2013 #17

    jedishrfu

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    Many times these errors creep into problem sets as they edit the problem to make it tougher. Looking at it though I couldn't quite reconstruct what the earlier problem wanted proven. It may have been given z= zbar to demonstrate b=0.
     
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