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Homework Help: Limit of a complex function

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Used the definition of a limit to prove that as z=>0 lim (z bar)^2/(z)=0

    2. Relevant equations

    abs(f(z)-w(0)) < eplison whenever abs(z-z(0)) < lower case delta

    3. The attempt at a solution

    let z=x+iy and z bar = x-iy


    Since limit of function is approaches origin, there are two cases when the limit approaches the origin: when (x,0) and when (y,0)

    first case(real axis): z=>(x,0) lim (z bar)^2/(z)=0 => (x-i*0)^2/(x+i*0)=(x^2)/x= x

    second case(imaginary axis) : z=>(0,y) lim (z bar)^2/(z)=0 => (0-i*y)^2/(0+i*y)=(i*y)^2/(i*y)= i*y

    Both cases each time show that the function has two different limits, and that the limit of the function does not approach zero in either case.

    So how can the limit of the function be zero?
  2. jcsd
  3. Sep 1, 2008 #2


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    I can't understand what you're doing. What does "when (x,0) and when (y,0)" mean?

    Also, you should have [itex]0 < |z - z_0| < \delta[/itex] in the definition of a limit; the "0 <" part is important.

    Now to solve your problem, I recommend that you first prove the following basic facts:
    (1) |z/w| = |z|/|w| (w[itex]\neq[/itex]0)
    (2) |[itex]\overline{z}[/itex]| = |z|.
  4. Sep 1, 2008 #3
    So I have to show that |z_bar^2/z|=|z_bar*z_bar/z|=|z_bar|*|z_bar|/z
  5. Sep 1, 2008 #4


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    I was hoping you'd see that, for z[itex]\neq[/itex]0,

    [tex]\left| \frac{\bar{z}^2}{z} \right| = |z|.[/tex]
  6. Sep 1, 2008 #5
    but wouldn't |z_bar^2/z| => |(x-iy)^2/(x+iy)| since z_bar is the conjugate of z. So how can |z_bar^2/z|=|z|?
  7. Sep 2, 2008 #6


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    ?? If z is approaching 0, both x and y go to 0. Those certainly do go to zero!
  8. Jul 15, 2009 #7
    please see attachment

    Attached Files:

  9. Jul 15, 2009 #8
    [tex]$\left\vert \frac{\bar{z}^2}{z} - 0 \right\vert = \left\vert
    \frac{\bar{z}^2}{z} \right\vert = |\bar{z} \bar{z} z^{-1}| =
    |\bar{z}| |\bar{z}| |z^{-1}| = |z||z||z^{-1}| = |z||z z^{-1}| = |z|
    < \varepsilon$[/tex]
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