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

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  • #1
s3a
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Homework Statement


The problem along with its solution are attached as TheProblemAndSolution.jpg. This post's focus is on part (iii), specifically.

Homework Equations


z (overlined) = z. (according to book)
z (double overlined) = z. (according to me)

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!
 

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Answers and Replies

  • #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.
 
  • #3
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I can't see your jpg TheProblemAndSolution.jpg

Can you provide a link for it or upload it?
 
  • #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.
 
  • #5
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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
 
  • #6
Dick
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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.
By z is real, I just mean b=0 in z=a+bi.
 
  • #7
s3a
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Sorry for the triple post.

Look two posts down.
 
  • #8
s3a
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Sorry for the triple post.

Look one post down.
 
  • #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. :)
 
  • #10
Dick
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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. :)
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.
 
  • #11
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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.
Yes, it was named in honor of Bar Rafaeli:

http://en.wikipedia.org/wiki/Bar_Refaeli
 
  • #12
Dick
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  • #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. :)
 
  • #14
Dick
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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. :)
Sure, joke. You're welcome!
 
  • #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.
 
  • #16
Dick
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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.
You're right. I didn't even look at the 'proof'. Pretty incoherent. I'm sure s3a will do better.
 
  • #17
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You're right. I didn't even look at the 'proof'. Pretty incoherent. I'm sure s3a will do better.
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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