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Complex analysis

  1. Jun 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Let f(z) = sqrt(z) be the branch of the square root function with sqrt(z) = (r^1/2) (e^iΘ/2),
    0≤Θ<2[itex]\pi[/itex], r > 0

    (a) for what values of z is sqrt(z^2) = z?

    (b) Which part of the complex plane stretches, and which part shrinks under this transformation?

    2. Relevant equations



    3. The attempt at a solution

    Ok so for this branch i believe the function will map all points within 0≤Θ<2[itex]\pi[/itex] to the upper half plane (ie Im(f) > 0).

    I do not really understand what part a is asking and for part b it seems everything is shrunk.
     
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  3. Jun 7, 2014 #2

    Simon Bridge

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    For part (a), it is asking for ##z:f(z^2)=z##
    For part (b) please show your reasoning. What does it mean to say that the complex plane has shrunk or stretched? How would you tell?
     
  4. Jun 7, 2014 #3
    For part a do we have to consider the principal nth root? Or particular branches of the square root function?

    In otherwords where this function is not multivalued?
     
  5. Jun 7, 2014 #4

    micromass

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    No, the function is not multivalued. Your OP has given the branch which you should consider:

    [tex]f(re^{i\theta}) = \sqrt{r} e^{i\theta/2}[/tex]

    where ##0\leq \theta< 2\pi##. That last restriction on ##\theta## makes sure it's not multivalued.
     
  6. Jun 7, 2014 #5
    then wouldnt part a be true for all z then? It seems obvious that sqrt(z^2) = z for all z. Why would that not be the case?
     
  7. Jun 7, 2014 #6

    micromass

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    Can you prove it?
     
  8. Jun 7, 2014 #7

    pasmith

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    Let [itex]z = e^{i3\pi/2}[/itex]. Then
    [tex]z^2 = (e^{i3\pi/2})^2 = e^{i3\pi} = e^{i\pi}[/tex] since we need [itex]0 \leq \arg(z^2) < 2\pi[/itex]. But then
    [tex]f(z^2) = (e^{i\pi})^{1/2} = e^{i\pi/2} \neq z.[/tex]
    That's one [itex]z[/itex] for which [itex]f(z^2) \neq z[/itex]. Are there others?
     
  9. Jun 7, 2014 #8
    It seems the equation fails for values of z where when you square them the argument is outside of 0 to 2pi
     
  10. Jun 7, 2014 #9

    lurflurf

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    ^Good which z are those? What can you say about their real and imaginary parts?

    For (b) Suppose we have two nearby points so that d(P1,P2)=h
    what can we say about d(sqrt(P1),sqrt(P2))?
    which is bigger? What is the formula for distance? (you could also consider areas)
     
  11. Jun 7, 2014 #10
    For part a it would be all z such that 0≤arg(z)≤pi. Therefore im(z) > 0.

    Is the distance formula just the standard Euklidian distance formula?
     
  12. Jun 8, 2014 #11

    lurflurf

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    ^yes consider a short segment in the complex plane
    If the length is h say what is the length after taking square root
    Where in the plane does a segment stretch and where does it shrink?
    Hint find where the length does not change. That region will separate the other two.
     
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