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Complex number epsilon-delta problem

  1. Feb 23, 2015 #1

    jjr

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    1. The problem statement, all variables and given/known data
    Prove that [itex] \lim_{z\rightarrow z_0} Re\hspace{1mm}z = Re\hspace{1mm} z_0 [/itex]

    2. Relevant equations
    It is specifically mentioned in the text that the epsilon-delta relation should be used,

    [itex] |f(z)-\omega_0| < \epsilon\hspace{3mm}\text{whenever}\hspace{3mm}0<|z-z_0|<\delta [/itex].

    Where [itex] \lim_{z\rightarrow z_0}f(z) = \omega_0 [/itex]
    Other equations that might be useful are
    [itex] |z_1+z_2| \leq |z_1| + |z_2|\hspace{3mm}\text{(Triangle inequality)} [/itex]
    and perhaps
    [itex] Re\hspace{1mm}z \leq |Re\hspace{1mm} z| \leq |z| [/itex]
    3. The attempt at a solution
    Here [itex] \omega_0 = Re\hspace{1mm}z_0[/itex] and [itex] f(z) = Re\hspace{1mm} z [/itex],
    so we want to find a delta [itex] |z-z_0| < \delta [/itex] such that [itex] |Re\hspace{1mm}z - Re\hspace{1mm}z_0| < \epsilon [/itex].

    I am honestly not sure how to approach this. Any clues would be very helpful.

    Thanks,
    J
     
  2. jcsd
  3. Feb 23, 2015 #2

    Svein

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    Start with the fact that limz→z0z = z0.
     
  4. Feb 23, 2015 #3

    Dick

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    Think about it. Which is larger ##|z-z_0|## or ##|Re(z)-Re(z_0)|##?
     
  5. Feb 23, 2015 #4

    jjr

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    Well, the distance between two complex points is definitely larger or equal to the distance between the real parts of the same points...
     
  6. Feb 23, 2015 #5

    Dick

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    Ok, so how small shall we make ##|z-z_0|## to make sure that ##|Re(z)-Re(z_0)| \lt \epsilon##?
     
  7. Feb 23, 2015 #6

    jjr

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    Hmm. I suppose that if we set [itex] |z-z_0| [/itex] smaller than [itex] \epsilon [/itex], then [itex] |Re(z)-Re(z_0)| [/itex] would have to be smaller than [itex] \epsilon [/itex]. I.e. [itex] \delta = \epsilon[/itex]?
     
  8. Feb 23, 2015 #7

    Dick

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    Yes.
     
  9. Feb 23, 2015 #8

    jjr

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    Thank you. Not too hard to see really, but I'm still not sure how to show it. Is the statement I made sufficient? I feel like it's lacking some rigor
     
  10. Feb 23, 2015 #9

    Dick

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    You just need to show ##|Re(z)| \le |z|##.
     
  11. Feb 24, 2015 #10

    jjr

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    Okay, I think I've got my head wrapped around it. I have another one, if you would care to have a look:

    To be shown:
    ##\lim_{z\rightarrow z_0} \bar{z}=\bar{z_0}##

    So
    ## |\bar{z}-\bar{z_0}|<\epsilon \to |z-z_0|<\delta##

    My work:
    I'm trying to find some relation between ##z,z_0## and their conjugates, so I tried setting
    ##\bar{z}=z-2Im(z)##
    ##\to |\bar{z}-\bar{z_0}| = |z-z_0-2Im(z)+2Im(z)|##.
    Here I tried invoking the triangle inequality, but the expression I ended up with didn't help much.
    Suggestions?

    Thanks,
    J

    edit: typo
     
    Last edited: Feb 24, 2015
  12. Feb 24, 2015 #11

    Svein

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    Hint: [itex]\lvert z-z_{0}\rvert ^{2}=(z-z_{0})(\bar{z}-\bar{z_{0}})=\lvert \bar{z}-\bar{z_{0}}\rvert ^{2} [/itex]
     
  13. Feb 24, 2015 #12

    jjr

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    Thank you. I suppose one could take the square root from both sides and obtain ##|z-z_0| = |\bar{z}-\bar{z_0}|##, which in turn enables us to set ##\delta = \epsilon##? Thinking about this again I realize that this makes sense, that the distance between two points ##z, z_0## should be the same as the distance between the same two points mirrored about the imaginary axis.
     
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