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Demonstration for Re(z), Im(z), Abs(z) and Arg(z)

  1. Jan 28, 2014 #1
    Someone can demonstrate me why

    ##Re(z) = \frac{1}{2} \left ( z+\bar{z} \right )##
    ##Im(z) = \frac{1}{2i} \left ( z-\bar{z} \right )##
    ##Abs(z)=\sqrt{z\bar{z}}##
    ##Arg(z)=-i ln\left ( \frac{z}{\sqrt{z\bar{z}}} \right )##

    ?

    2# Is correct to affirm that

    ##Arg(z)=-i ln\left (\sqrt{\frac{z}{\bar{z}}} \right)##

    ?
     
  2. jcsd
  3. Jan 28, 2014 #2

    jbunniii

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    Try writing ##z## as either ##x + iy## or ##r e^{i\theta}##, whichever is most suitable in each case. If you get stuck somewhere, please show what you tried.
     
  4. Jan 29, 2014 #3

    Ray Vickson

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    PF Rules require you to show your work. We do not do homework here; we just give hints and suggestions.
     
  5. Jan 29, 2014 #4
    homework? no comments...

    I'm asking for a demonstration. Demonstration in math is a serious thing.
     
  6. Jan 30, 2014 #5

    Mark44

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    That's not relevant. If the question is about homework or textbook problems, the rules here require that you show what you have tried.
     
  7. Jan 30, 2014 #6

    Mentallic

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    I find it odd that the OP is asking for proof for the representation of Arg(z) as well as Re(z). The difference in difficulty between the two of these is quite profound.


    Start with answering these questions:
    What is the standard representation for z, a complex number? Hint: It's already been said in this thread.
    What is Re(z) equal to? What about Im(z)?
    What is [itex]\bar{z}[/itex]?
     
  8. Jan 30, 2014 #7
    I already show my hypotheses a lot of times in others topics but, in general, the answer that I have received are, nearly always, a specie of subterfuge. If someone ask how much is 2+2 the answers are (in general) "the sum was the first discovery of man...", "the equality is reflexive, replacement, transitive, symmetric..." etc,etc,etc. But the answer 2+2 is equal to 2 not is given.
     
  9. Jan 30, 2014 #8

    Mark44

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    No wonder, because 2 + 2 ≠ 2.

    Since you have refused to show any sort of effort on this problem, I am closing this thread.
     
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